Calculate derivative of erfcx(x) function

I want to know how i can calculater the derivative of an erfcx(x) function with the symbolic math toolbox. The following is what i tried so far:

>> syms x real
>> y = erfcx(x)
Undefined function 'erfcx' for input arguments of type 'sym'.

I also tried doing this with an anonymous function:

>> testing = @(x) erfcx(x)

testing =

  function_handle with value:

    @(x)erfcx(x)

but then i don't know how to use the diff() function, as i don't know what variable i should give: diff(testing(x)) <- instead of this x.

[madhan ravi]

I suspect there is no analytical derivative for erfcx(x) therefore why not approximate it numerically?

[Kevin Frey]

do you have a link/doi for me where this is described?

[Bjorn Gustavsson]

HUH? No analytical derivative of erfcx?

Try:

help erfcx
help erfc
help erf

Then apply the definition of erf in producing a version of erfcx that the symbolic toolbox can handle.

[madhan ravi]

Without transforming erfcx() is it possible to differentiate it?

[Kevin Frey]

so the base problem with inserting "erfcx(x) = exp(x^2) * erfc(x)" before calculating the derivative (which i already did via wolfram alpha) is that i get overflow errors and i wanted to circumvent that issue with the erfcx(x) function.

So if that is what you two are suggesting i am afraid that won't work for me.

But i really appreciate all your input.

[Bjorn Gustavsson]

Why cant you do that and plug it into the symbolic differentiation? (It took me ~5 minutes to get an analytical derivative...)

Answer 1

[Torsten]

Try

x = 0:0.1:5;
analytical_derivative = -2/sqrt(pi) + 2.*x.*erfcx(x);
numerical_derivative = (erfcx(x+1.0e-6)-erfcx(x))*1e6;
plot(x,analytical_derivative,x,numerical_derivative)

So my guess is that the derivative of erfcx is erfcx'(x) = -2/sqrt(pi) + 2*x*erfcx(x) (and you can easily deduce this result by differentiating exp(x^2)*erfcx(x) using the product rule).

Answer 2

[Bjorn Gustavsson]

Just plug in the definitions of the erfcx function into your symbolic calculations? Or do the differentiation by hand?

HTH

[Kevin Frey]

it was mentioned that you should use the erfcx instead of e(t^2)*erfc(t) to avoid overflow.

"Description

erfcx(x) returns the value of the Scaled Complementary Error Function for each element of x. Use the erfcx function to replace expressions containing exp(x^2)*erfc(x) to avoid underflow or overflow errors."

Thats why i wanted to use that function, or am i missing something?

[Steven Lord]

If you're performing the calculations symbolically (which you will need to do to take the derivative using diff) this isn't a problem. Just don't try to evaluate your result for a large numeric value (or if you do, convert that number into a sym first so you perform the calculation using variable precision arithmetic.)

[Bjorn Gustavsson]

Perhaps. You asked for an analytical derivative in the original post. For symbolic calculations you will not have problems with numerical underflow nor overflow. So I assumed that something like an anlytical derivative would be what you wanted for some purpose or the other, and suggested a simpe 2-3-step way to get you that. If you want a numerical derivative then simly use the gradient function for that.

[Kevin Frey]

I am sorry, i have problems understanding what exactly you mean by numerical or analytical. From what i found "analytic" and "symbolic" are supposed to be the same and they are meant to be exact and "numerical" are approximations.

So what i need in the end, is a working function for a exponentially modified gauss calculcation i am currently working on, and if i use the approximation of "erfcx(x) = exp(x^2) * erfc(x)" i get overflow results, if my number gets too big and i wanted to circumvent that.

[Bjorn Gustavsson]

If I got the definitions right you should just simply have expressed erfcx in terms of exp(x^2) and erf:

syms x
diff(exp(x^2)*(1-erf(x)),x)
ans =
- 2/pi^(1/2) - 2*x*exp(x^2)*(erf(x) - 1)

HTH