Hi

Being a holiday here, I had some extra time to play with this instrument.

Here is an issue with how the prime handles derivatives of functions that I do not comprehend. Maybe there is documentation in xcas about it, but ....

I define a function

b(x):=IFTE(x=0,1,x/(exp(x)-1)

This is a nice function that has a power series around 0 whose coefficients are the Bernoulli number B(n) divided by n!

So the derivative b'(0)=-1/2, but the HP prime computes b'(0)=+/- inf.

I cannot figure out how it does that. Thanks!

*Edited: 28 Nov 2013, 12:58 p.m. *

What happens if you use the more exact and dedicated *e*^{x}-1 command in the denominator, i.e. EXPM1(x) ?

Dieter

*Edited: 28 Nov 2013, 1:09 p.m. *

Thanks, I did not know about EXPM1. But if I set

h(x):=IFTE(x=0,1,x/(EXPM1(x)-1))

I obtain (if not mistaken)

h'(0)=-1

Alberto, EXPM1(x) already subtracts 1. That's why it says M1 (minus one). You must not add another subtraction. ;-)

h(x):=IFTE(x=0,1,x/EXPM1(x))

Compare the results of EXP(x) - 1 with those obtained by using EXPM1(x) for some x close to zero, and you will see why this special command exists.

Dieter

*Edited: 28 Nov 2013, 1:40 p.m. *

Ah! Thank you for the clarification.

But then

h(x):=IFTE(x=0,1,x/EXPM1(x))

results in

h'(0)=+/- inf

It may be something with xcas algorithms for derivatives.

Dunno.

However, you can natively do piecewise functions. I'd use that instead. I suspect that will be easier.

TW

*Edited: 28 Nov 2013, 2:15 p.m. *

There is a bug that I'm fixing in the source code, infinity-undef should return undef, not infinity, hence you should have undef as result.

You must use limit to find the derivative at a singular point, if you simply evaluate, then x is replaced by 0 in the analytic expression of the derivative (which does not take care of an ifte defined only for one point).

OK, thanks.

BTW, what is the real difference between exp(x)-1 and EXPM1(x)?

The example given in the Help of the prime is EXPM1(0.23) and this is the same as exp(0.23)-1 for the Prime. But even much smaller numbers like x=0.00001 lead to the same result.

EXPM1(x) is far more accurate as x approaches zero. Consider a small x-value like 1E-12. With 12-digit accuracy, EXP(x) is evaluated as excactly 1, so EXP(x)-1 becomes zero due to roundoff.

Here EXPM1(x) comes to rescue. It is evaluated differently so that the result is exact. Here are some results for 12 digit working precision:

x EXP(x) EXP(x)-1 EXPM1(x)

---------------------------------------------------------------

0,23 1,25860000993 0,25860000993 0,258600009929

0,0022 1,00220242178 2,20242178000 E-3 2,20242177564 E-3

1 E-5 1,00001000005 1,00000500000 E-5 1,00000500002 E-5

1/9 E-7 1,00000001111 1,11110000000 E-8 1,11111111728 E-8

1/7 E-9 1,00000000014 1,40000000000 E-10 1,42857142867 E-10

1/6 E-11 1,00000000000 0,00000000000 1,66666666667 E-12

So EXPM1(x) (and also its inverse function LNP1(x) = ln(1+x)) allow exact results where their conventional counterparts would return inexact or even useless results. There are extremely useful in many everyday applications.

Dieter

*Edited: 29 Nov 2013, 3:17 p.m. *

Thank you again, Dieter. The example EXPM1(0.23) given in Help is very misleading. Are there any references to how these functions are evaluated?