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From: Ilyas Yildirim <yildirim.pas.rochester.edu>

Date: Tue, 8 Mar 2005 15:56:48 -0500 (EST)

The main function I am interested to use is tanh; therefore did not

write the exact form of the function. Sorry for that. Here is the function

I am interested to use:

f(x,k) = {1+tanh[k*(2x-1)]}/2

where x = lambda.

=> V(lambda) = f(lambda,k) * V0 + [1 - f(lambda,k)] * V1

I am particularly interested on the k values 3,4,5,6. I am not sure what

you mean by "gradual". When I plot this function for the above k values,

they are both zero and one at lambda=0 and lambda=1, respectively. And

compared to (1-lambda)^k, I do not see a big difference, except that it is

zero at lambda=0. It seems like a gradual function, but as I said, I dont

know how gradual this function should be in order it to be used in TI

calculations.

Best regards,

On Tue, 8 Mar 2005, David A. Case wrote:

*> On Thu, Mar 03, 2005, Ilyas Yildirim wrote:
*

*>
*

*> > In my prev. emails, I was asking why AMBER does not let us use dummy
*

*> > atoms in the initial state when k is not equal to 1 (pp.134, note #2). And
*

*> > as far as I understand, the reason lies at the parameter function chosen
*

*> > in Eq. # 5 (pp.133). When I plot the function, (1-lambda)^k, it seems
*

*> > that when we have dummy atoms at the initial state, the transformation
*

*> > will be sharp. It is not as gradual as it is when we have dummy atoms at
*

*> > the final state.
*

*> >
*

*> > So, we were thinking that if we use a parameter function,
*

*> > tanh[k(2lambda-1)], such that the Eq. 5 becomes,
*

*> >
*

*> > V(lambda)= tanh[k(2lambda-1)]*V0 + {1 - tanh[k(2lambda-1)]}*V1
*

*> >
*

*> > the problem might be solved. When we plot the function tanh[k(2lambda-1)],
*

*> > we see that the initial and the final transformations look gradual.
*

*>
*

*> Just a quick note: the function you use not only has to "look gradual", you
*

*> have to be able to show (presumably analytically) that it decays very stongly
*

*> as a function of lambda, so that the net effect of vdW interactions remains
*

*> finite. Offhand, it doesn't look like your function does this; the
*

*> coefficient of V0 doesn't even go to zero as lambda -> 0. Apologies if I am
*

*> missing something here....
*

*>
*

*> ....dac
*

*>
*

*> -----------------------------------------------------------------------
*

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*

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*

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*

*>
*

*>
*

Date: Tue, 8 Mar 2005 15:56:48 -0500 (EST)

The main function I am interested to use is tanh; therefore did not

write the exact form of the function. Sorry for that. Here is the function

I am interested to use:

f(x,k) = {1+tanh[k*(2x-1)]}/2

where x = lambda.

=> V(lambda) = f(lambda,k) * V0 + [1 - f(lambda,k)] * V1

I am particularly interested on the k values 3,4,5,6. I am not sure what

you mean by "gradual". When I plot this function for the above k values,

they are both zero and one at lambda=0 and lambda=1, respectively. And

compared to (1-lambda)^k, I do not see a big difference, except that it is

zero at lambda=0. It seems like a gradual function, but as I said, I dont

know how gradual this function should be in order it to be used in TI

calculations.

Best regards,

On Tue, 8 Mar 2005, David A. Case wrote:

-- Ilyas Yildirim --------------------------------------------------------------- - Department of Chemisty - - - University of Rochester - - - Hutchison Hall, # B10 - - - Rochester, NY 14627-0216 - Ph.:(585) 275 67 66 (Office) - - http://www.pas.rochester.edu/~yildirim/ - --------------------------------------------------------------- ----------------------------------------------------------------------- The AMBER Mail Reflector To post, send mail to amber.scripps.edu To unsubscribe, send "unsubscribe amber" to majordomo.scripps.eduReceived on Tue Mar 08 2005 - 21:53:00 PST

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