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Re: Fwd: RE: Antiderivatives and Definite Integrals
*To*: mathgroup at smc.vnet.net
*Subject*: [mg39874] Re: [mg39841] Fwd: [mg39713] RE: [mg39670] Antiderivatives and Definite Integrals
*From*: Dr Bob <drbob at bigfoot.com>
*Date*: Sun, 9 Mar 2003 05:29:53 -0500 (EST)
*References*: <200303080750.CAA23648@smc.vnet.net>
*Reply-to*: drbob at bigfoot.com
*Sender*: owner-wri-mathgroup at wolfram.com
Here's an odd result:
r = 2 - Sin[t];
f = Sqrt[r^2 + D[r, t]^2]
g = Simplify@f
Integrate[f, {t, 0, Pi/2}] // N
Integrate[g, {t, 0, Pi/2}] // N
NIntegrate[f, {t, 0, Pi/2}]
NIntegrate[g, {t, 0, Pi/2}]
-14.474963421625041 + 3.552713678800501*^-15*I
2.382540126500918 + 0.*I
2.3825401265009183
2.3825401265009183
(Three of those answers are correct.)
Simplifying f changed the result from Integrate, even though it doesn't
change the plot:
DisplayTogetherArray[Plot[f, {t, 0, Pi/2}],
Plot[g, {t, 0, Pi/2}], AspectRatio -> Automatic, ImageSize -> 500];
The moral, as always, is "caveat emptor".
Bobby
On Sat, 8 Mar 2003 02:50:48 -0500 (EST), Garry Helzer <gah at math.umd.edu>
wrote:
>
>
> Begin forwarded message:
>
>> From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
>> Date: Mon Mar 3, 2003 11:48:37 PM US/Eastern
>> To: mathgroup at smc.vnet.net
>> Subject: [mg39874] [mg39841] [mg39713] RE: [mg39670] Antiderivatives and Definite
>> Integrals
>>
>>
>>> -----Original Message-----
>>> From: Garry Helzer [mailto:gah at math.umd.edu]
To: mathgroup at smc.vnet.net
>> To: mathgroup at smc.vnet.net
>>> Sent: Saturday, March 01, 2003 8:48 AM
>>> To: mathgroup at smc.vnet.net
>>> Subject: [mg39874] [mg39841] [mg39713] [mg39670] Antiderivatives and Definite
>>> Integrals
>>>
>>>
>>> The antiderivative of Sqrt[1+Cos[x]] discussed here recently (sorry, I
>>> lost the thread) provides an amusing illustration of the fact that
>>> Mathematica does not always evaluate definite integrals by first
>>> finding an antiderivative and then substituting in the upper and lower
>>> limits. (See the Mathematica book A.9.5) Make the definitions
>>>
>>> f[x_] = Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>>> g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>>>
>>> Then f[2Pi] is 0 (wrong) and g[2Pi] if 4Sqrt[2] (correct). Of course
>>> f[x]==g[x] returns True.
>>>
>>> Garry Helzer
>>> Department of Mathematics
>>> University of Maryland
>>> 1303 Math Bldg
>>> College Park, MD 20742-4015
>>>
>>>
>>
>> You might be interested to observe:
>>
>>
>> In[1]:= f[x_] = Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>> Out[1]= 2*Sqrt[1 + Cos[x]]*Tan[x/2]
>>
>> In[2]:= g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>>
>> In[3]:= ?g
>> Global`g
>> g[x_] := Integrate[Sqrt[1 + Cos[t]], {t, 0, x}]
>>
>> In[4]:= Plot[f[x], {x, 0, 2Pi}] // Timing
>> Out[4]= {0.03 Second, = Graphics =}
>>
>> ...what you called "wrong".
>
> Well, it is wrong. From calculus we know that
> 1. A function defined as an integral with a variable upper limit is an
> antiderivative.
> 2. An antiderivative is, by definition, differentiable.
> 3. A differentiable function is continuous.
>
> So, by this plot, f[x] is not the antiderivative of any function.
>
> But this may be too rigid a view. After all, D[f[x],x]//FullSimplify
> brings you back to the integrand. This is a formal computation that
> misses the points of discontinuity. Perhaps the algorithms used by
> Mathematica are guaranteed only to produce such formal antiderivatives. A
> superficial look a the Risch algorithm suggests that this might be the
> case since it proceeds by formal manipulation in extension fields of
> fields of rational functions.
>
> As I was typing this a stranger wandered in looking for an arc length(it
> happens). His problem provides another example. Let r=2-Sin[t] and try
> Integrate[Sqrt[r^2 +D[r,t]^2],{t,0,Pi/2}]//N. The result is about -15,
> but arc length should not be negative. Set
>
> h[x_]= Integrate[Sqrt[r^2 +D[r,t]^2],{t,0,x}]
>
> and again we get a formal antiderivative , the plot of which reveals
> discontinuities. It cannot be carelessly used to evaluate definite
> integrals. (If you simplify Sqrt[r^2 +D[r,t]^2] before integrating you
> get an honest antiderivative.)
>
> In[5]:=
>> Plot[g[x], {x, 0, 2Pi}, PlotRange -> {All, {0, 6}}] // Timing
>> Out[5]= {4.096 Second, = Graphics =}
>>
>> ...what you called "right".
>
> It is right. But the corresponding function for the arc length problem is
> wrong.
>>
>>
>> Redefine:
>>
>> In[6]:= f[x_] /; -Pi < x < Pi = f[x]
>> Out[6]= 2*Sqrt[1 + Cos[x]]*Tan[x/2]
>>
>> In[7]:= f[x_] /; Pi < x < 3 Pi = f[x] + 4*Sqrt[2]
>> Out[7]= 4*Sqrt[2] + 2*Sqrt[1 + Cos[x]]*Tan[x/2]
>
> Or 4Sqrt[2]Round[x/(2Pi)]+ 2*Sqrt[1 + Cos[x]]*Tan[x/2] . But these
> formulas are less than perfect since they are indeterminate at odd
> multiples of Pi.
>
>>
>>
>> In[9]:=
>> Plot[f[x], {x, 0, 2Pi}, PlotRange -> {All, {0, 6}}] // Timing
>> Out[9]= {0.03 Second, = Graphics =}
>>
>> Don't be confused by that double use of f (as a function, and as a
>> shortcut
>> for an algebraic expression).
>>
>> --
>> Hartmut Wolf
>>
>>
>>
> Garry Helzer
> Department of Mathematics
> University of Maryland
> 1303 Math Bldg
> College Park, MD 20742-4015
>
>
>
--
majort at cox-internet.com
Bobby R. Treat
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