Math Problem

Using the multivariate second derivative test to classify the critical points of the function may help:

Are you looking for a closed solution or a "feel" for the domain? Just

so the full term will range over 0 to 1 times V or at most V.

The second term will always be non-negative when F and W are non-negative. The only time f(v) can be negative is if F is negative and the values of F and W (as well as V of course) are such that the magnitude of the second term is larger than the first. I can't pin that down in a closed form but I think this can help you intuitively map the domain a bit more.

Is that a big dot on top of f(v)? You do understand the dot is conventionally reserved for time derivatives.

Thanks! ...Jim Thompson

So, are V, K, W all real numbers? And the dot indicates derivative, so you want to know the function is nondecreasing (i.e. weakly monotonic)?

You're giving a function, not in terms of time.

Then you're asking a question about it's time derivative.

Either that or that very deliberate-looking dot above f(v) on the bottom line is a typo. (What's a typo called when something is hand-written?)

Or do you mean f'(v), i.e. d/dv f(v)?

Yes

Yes

It should not have a negative slope for any "V"

I have stumbled onto an identity that may help me...

SECH^2(u) = 1 - TANH^2(u) ...Jim Thompson

d/dv

Picky, picky ;-) ...Jim Thompson

Icky icky.

I flogged it a bit. Partial answers involve transcendental functions, so I suspect there is no closed-form solutions. If there are, they're not available without lots more flogging.

Ah, I'd played with this a little bit before.

Ultimately, I believe Wescott's right. Intuitively, you're emulating a resistor and diode circuit -- a system which SPICE solves iteratively, and a solution which is transcendental (you can only iterate the x[n] = e^x[n-1] form to get an approximate result, there's no closed form analytical solution -- see

Other sigmoid functions come to mind; you can use the Fermi-Dirac statistic (aka Logistic function), 1 / (1 + exp(-x)), to "turn on" the proportional part. But if you unwrap the tanh function, you'll see that's the same thing you're already doing (give or take some constants).

In principle, there exists a function that exactly fills in that little remainder, that you're trying to patch with the Gaussian term. It's most likely transcendental as well... and may not even have a simple description other than to being the remainder to the function in question!

Other sigmoids will have other shapes. Taking the integral of exp(-x^2) might help (erf(x)). There are other kinds out there. Do consider that, since you're SPICEing, you can do some quite excellent numerical solutions, and may not be able to express your desired relation as an equation, but you might be able to with reasonable numerical stability and few additional nodes.

Tim

f(v) = [ v + v.tanh(kv) ] /2 + Fexp(-wv^2) f'(v) = [1 + tanh(kv) + kv.sech^2(kv) ] /2 - 2Fwv.exp(-wv^2)

Which as you say using the identity above simplifies a bit

f'(v) = [1 + tanh(ky) + kv - kv.tanh^2(kv) ] /2 - 2Fwv.exp(-wv^2)

Sanity checks v= -inf f' = 0 f=0 v = 0 f' = 1/2 f~[v + kv^2]/2 + F(1-wv^2) v = inf f' = 1 f=v

The negative gradient terms are kv.tanh^2(kv) and -2Fwv.exp(-wv^2)

They are each worst case when their next derivative is zero ie.

ktanh^2(kv) + 2k^2vtanh^3(kv)(1- tanh^2(kv) = 0

hence a rough constraint on k alone for worst case v subst x = tanh(kv)

kx^2 + 2k^2vx^3 - 2k^2vx^4 = 0

trivial solutions x= 0, k=0 uninteresting or

x = -kv +/- sqrt((kv)^2 + 2kv)

And then for the Gaussian

(-2Fw + 4Fw^2v^2).exp(-wv^2)

hence v^2 = 1/2w (classic result for a Gaussian)

Not sure you can obtain a nice closed form solution for the valid bounds on k,F,w but it is clear that you should be worried if |Fw| is large or |k| is small (all subject to typos and algebra slips)

You may be able to combine them on a both components at worst case assumption to rule out a large chunk of parameter space.

It might be a little bit more amenable to an analytic closed form solution if he was prepared to substitute F/(1+wv^2) for F.exp(-wv^2)

My bad... in my curve-fitting the error looked like (Gaussian) exp(-W*V^2)

Looks the same but easier to deal with: 1-[TANH(K*V)]^2 ...Jim Thompson

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