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Re: Erf funcion

  • To: mathgroup at smc.vnet.net
  • Subject: [mg123956] Re: Erf funcion
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Mon, 2 Jan 2012 02:46:16 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201201010729.CAA28444@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

Oops yet again. Gotta stop hitting send so fast!

The rule does make a difference:

Clear[FiX]
rule = Erf[x_] :> 2*FiX[x*Sqrt[2]] - 1;
expr = 3 (2^(7/2 -
       2 (k + t))) Cp (E^(-d^2 - (36 Cp^2)/
        lambda^2)) ((Erf[(-d + (6 Cp)/lambda)/Sqrt[2]] +
       Erf[(6 Cp + d lambda)/(Sqrt[2] lambda)])^k)
expr /. rule // Simplify

3 2^(7/2 - 2 (k + t)) Cp E^(-d^2 - (36 Cp^2)/
   lambda^2) (Erf[(-d + (6 Cp)/lambda)/Sqrt[2]] +
    Erf[(6 Cp + d lambda)/(Sqrt[2] lambda)])^k

3 2^(7/2 - k - 2 t) Cp E^(-d^2 - (36 Cp^2)/
   lambda^2) (-1 + FiX[-d + (6 Cp)/lambda] + FiX[d + (6 Cp)/lambda])^k

I'm not sure that's as simple as you'd like, however.

Bobby

On Sun, 01 Jan 2012 18:30:58 -0600, DrMajorBob <btreat1 at austin.rr.com>  
wrote:

> You can eliminate Erf this way:
>
> Clear[FiX]
> rule = Erf[x_] :> 2*FiX[x*Sqrt[2]] - 1;
> expr = 3 (2^(7/2 - 2 (k + t))) Cp (E^(-d^2 - (36 Cp^2)/lambda^2))
> ((Erf[(-d + (6 Cp)/lambda)/Sqrt[2]] +
>       Erf[(6 Cp + d lambda)/(Sqrt[2] lambda)])^k);
> expr /. rule // Simplify
>
> 3 2^(7/2 - 2 (k + t)) Cp E^(-d^2 - (36 Cp^2)/lambda^2)
>
> 3 2^(7/2 - 2 (k + t)) Cp E^(-d^2 - (36 Cp^2)/lambda^2)
>
> but as you can see, for this expression it makes no difference.
>
> Are you also trying to eliminate the exponential function?
>
> Bobby
>
> On Sun, 01 Jan 2012 01:29:27 -0600, Vicent <vginer at gmail.com> wrote:
>
>> Hello.
>>
>> I have an expression that depends on the Cumulative Distribution
>> Function (CDF) of a Normal Variable with mean = 0 and standard
>> deviation = 1.
>>
>> When I perform operations and/or derivatives with/for that function
>> with Mathematica, I get expressions involving Erf function, as the CDF
>> function of a Normal distributed variable and Erf (error function) are
>> closely related. Concretely,
>>
>>     FiX[x] == (1/2)*(1 + Erf[x/Sqrt[2]])
>>
>> and thus
>>
>>     Erf[x] == 2*FiX[x*Sqrt[2]] - 1
>>
>> where FiX stands for the previously mentioned CDF, I mean:
>>
>>     X = NormalDistribution[0, 1]
>>     FiX[x_] := CDF[X, x]
>>
>>
>> Is there a quick way to ask Mathematica not to use  Erf  but always
>> let the results as a function of FI[x]??
>>
>> The expressions I get as a result of my computations are larger than
>> this short example:
>>
>>     3 (2^(7/2 - 2 (k + t))) Cp (  E^(-d^2 - (36 Cp^2)/lambda^2))
>> ((Erf[(-d + (6 Cp)/lambda)/Sqrt[2]] + Erf[(6 Cp + d lambda)/(Sqrt[2]
>> lambda)])^k)
>>
>> where "Cp", "k", "t", "d" and "lambda" are parameters or arguments
>> that I want to keep as "generic".
>>
>> So, how can I "translate" expressions such as the last one into "FiX
>> mode"? Or, better, how can I prevent Mathematica to reduce everything
>> to "Erf mode"??
>>
>> I hope my question is clear enough... :)
>>
>> Thank you very much in advance for your answers.
>>
>> And have a good new year!
>>
>>
>> --
>> vicent
>> dooid.com/vicent
>>
>
>


-- 
DrMajorBob at yahoo.com



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