| Author |
Comment/Response |
jf
|
 06/23/12 6:39pm
In Response To 'Re: Re: Re: Re: Re: Limit returns unevaluated i...'
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Is your Assuming covering enough of the expression?
In[2]:= soln=Assuming[{b0>b1,b1>0,m0>m1,x\[Element]Reals&&n\[Element]Integers},Simplify[Limit[(1/Sqrt[2 \[Pi]])((E^(-(((b0^2-b1^2) (m0-m1) (-b1^2 (m0+m1 (-1+n)+3 m0 n-4 n x)+b0^2 (m0-m1+m0 n+3 m1 n-4 n x))-2 n (b1^2 (m0-x)+b0^2 (-m1+x))^2 PolyGamma[0,(b0^2+b1^2 (-1+n))/(b0^2-b1^2)]+2 n (b1^2 (m0-x)+b0^2 (-m1+x))^2 PolyGamma[0,(-b1^2+b0^2 (1+n))/(b0^2-b1^2)])/(4 (b0-b1)^3 (b0+b1)^3)))/(Sqrt[((b0^2-b1^2)/n)^n Pochhammer[1+(b1^2 n)/(b0^2-b1^2),n]]))^((1/n))),n->Infinity]]]
Out[2]=
(1/Sqrt[2*Pi])*((b1^(b1^2/(b0^2 - b1^2))*
E^((2*b0^4 + b1^2*(2*b1^2 + (m0 - m1)*(3*m0 + m1 - 4*x)) -
b0^2*(4*b1^2 + (m0 - m1)*(m0 + 3*m1 - 4*x)))/(4*(b0 - b1)^2*(b0 + b1)^2)))/
(b0^(b0^2/(b0^2 - b1^2))*(b0/b1)^((b1^2*(m0 - x) + b0^2*(-m1 + x))^2/
((b0 - b1)^3*(b0 + b1)^3))))
For sample values of these parameters,
In[3]:= soln /. {b1 ->1, b0->3, m0->2,m1->0}
Out[3]=
(3^(-(9/8) - (1/512)*(2 + 8*x)^2)*
E^((1/256)*(164 - 9*(4 + 2*(2 - 4*x)) + 2*(6 - 4*x))))/Sqrt[2*Pi]
Try a few values of x.
In[4]:= % /. {{x->0.},{x->1.},{x->10.}}
Out[4]=
{0.172525, 0.180288, 1.14914 10^-6 }
(Oops- two of those x's do not meet the assumption "x is between m0 and m1".)
In[5]:= $Version
Out[5]= 8.0 for Mac OS X x86 (64-bit) (October 5, 2011)
Attachment: sf29120.nb, URL: , |
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