Re: Re: Limit[Sin[a*x]/(a*x), x -> Infinity]
- To: mathgroup at smc.vnet.net
- Subject: [mg35840] Re: [mg35818] Re: Limit[Sin[a*x]/(a*x), x -> Infinity]
- From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
- Date: Sat, 3 Aug 2002 00:16:26 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
On Friday, August 2, 2002, at 03:42 PM, Dana DeLouis wrote:
>
> Hello. I do not have an answer, as I am new also.
> In Mathematica 4.2, Help says in part "...Limit therefore makes no
> explicit assumptions about symbolic functions." Same for the Package
> "NLimit." Therefore, I think we're out of luck here.
You can't make assumptions in Limit. There are not enough general
methods for such a mechanism to be really useful.
> As an newbe observation, the only option for Limit (For Mathematica 4.2)
> is "Direction." However, if you type ??Limit, there is another Option
> listed that appears not to be documented. That option is "Analytic." I
> thought maybe this would solve this type of problem, but it does not. I
> do not know for sure what this option does. Anyone else??
This is meant for the situations when you use a custom defined function.
For example:
In[21]:=
Limit[f[Sin[x]/x], x -> 0]
Out[21]=
Limit[f[Sin[x]/x], x -> 0]
but
In[22]:=
Limit[f[Sin[x]/x], x -> 0, Analytic -> True]
Out[22]=
f[1]
>
> Information["Limit", LongForm -> True]
> "Limit[expr, x->x0] finds the limiting value of \
> expr when x approaches x0...
>
> Attributes[Limit] = {Listable, Protected}
>
> Options[Limit] = {Analytic -> False, Direction -> Automatic}
>
> As a possible workaround, would substituting many values for "a" work?
> Here is what I am thinking...
>
> v = Table[Random[Real, {0, 2*Pi}], {100}];
>
> Union[(Limit[Sin[#1*x]/(#1*x), x -> Infinity] & ) /@ v]
>
> I get all zero's like you predicted for many values of a.
> {0}
>
> If you try "Integers", then I get an "Indeterminate" because the limit
> does not like "a" being 0.
>
> v = Table[Random[Integer, {-4, 4}], {10}];
>
> Union[(Limit[Sin[#1*x]/(#1*x), x -> Infinity] & ) /@ v]
>
>
> {0, Indeterminate}
Actually the problem is that the function Sin[z]/z is singular at
Infinity. Thus:
In[26]:=
Limit[Sin[z]/z, z -> Infinity, Direction -> -1]
Out[26]=
0
In[27]:=
Limit[Sin[z]/z, z -> Infinity, Direction -> -I]
Out[27]=
Infinity
In other words the limit depends on the direction. So when you consider
Limit[Sin[a z]/(a z), z -> Infinity]
it is not clear in which direction the limit is being taken. This
depends on whether a is real or not (as Bobby Treat already pointed this
out in his message),and of course there is the special case a==0.
Since Limit does nto allow assumptions to be entered if you know that
you will be dealing with such cases frequently one approach is to
redefine limit:
Unprotect[Limit];
In[28]:=
Limit[Sin[a_*x_]/(a_*x_), x_ -> Infinity] /;
Im[a] == 0 && a != 0 = 0;
Limit[Sin[a_*x_]/(a_*x_), x_ -> Infinity] /;
Re[a] == 0 && a != 0 = Infinity;
In[30]:=
Protect[Limit];
Now you need to enter assumptions about a's or b's, e.g.:
In[31]:=
a/:Im[a]=0;
In[32]:=
a/:(aâ? 0)=True;
this says a is a non-zero real. Limit will now return the expected
answer:
In[33]:=
Limit[Sin[a*x]/(a*x), x -> Infinity]
Out[33]=
0
similarly for the case when b is non zero and imaginary we will get:
In[34]:=
b /: Re[b] = 0;
In[35]:=
b /: b != 0 = True;
In[36]:=
Limit[Sin[b*x]/(b*x), x -> Infinity]
Out[12]=
Infinity
If you are sure you are going to onl be dealing with non-zero real
numbers you can redefine Limit to return always 0 in such cases, which
may make it simpler to deal with a large number of them in one go.
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
>
> --
> Dana
> Windows XP & Mathematica 4.1
> = = = = = = = = = = = = = = = = =
>
> "JM" <j_m_1967 at hotmail.com> wrote in message
> news:<aiat23$5b0$1 at smc.vnet.net>...
>> Sorry for 'refreshing' this message but does anyone know if I can
>> define the assumption below.
>>
>> I don't want to write specific assumptions for each term in Limit
>> since I have many different variables and forms of a. Is it possible
>> to generalise the assumption?
>>
>> it would really help me (and I really don't want to have to buy any
>> additional maths software).
>>
>>
>> timreh719 at yahoo.com.tw (bryan) wrote in message
>> news:<ahdf97$i36$1 at smc.vnet.net>...
>>> Hi All :
>>> I am also interesting in the solution of how to make an asumption
>
>>> in Mathematica. I can't find any method in Mathematica. If anybody
>>> has the approch to do this , please send it to my e-mail too , Thank
>
>>> you all ..
>>>
>>> j_m_1967 at hotmail.com (JM) wrote in message
>>> news:<ah8otr$aut$1 at smc.vnet.net>...
>>>> I know that this should be 0 but why can't I get mathematica to
>>>> think likewise.
>>>>
>>>>
>>>> In[4]:= Limit[Sin[a*x]/(a*x),x->Infinity]
>>>>
>>>> Sin[a x]
>>>> Out[4]= Limit[--------, x -> Infinity]
>>>> a x
>>>>
>>>> Is the problem a? How can I specify the properties of or
>>>> assumptions that may be made about a?
>>
>
>
>