Hi,
here my method for the exponentianal function e^x .
it is a mix of 2 fundamentals:
- Taylor series running at a 10 steps for decimals -1.0 < x < +1.0
- a couple of basic rules of the logarithms.

Explanation.
say we want to calculate e^20.3. well,
`` 20.3 float double is 40344CCCCCCCCCCDh hexadecimal``

we convert it to base2 by multiplying it by lg2(e)
``20.3 x 1.4426950408889632824453128103079fresult 20.3 base2 403D4965C85C0166h``

it is needed to know later how much we should shift left
our partial result in its integer part.
now, for the fundamental of powers we know that
``  n^20.3 can be rewritten as n^20 * n^0.3``

also we round down 20.3 base2 to get
`` 20.0 and 0.3 float double remainder  20.0 = 403D000000000000h  0.3  = 3FD2597217005980h``

we store the remainder 0.3 now for later use.

then we convert the integral part 20.0 base2 to an integer value,

20.0 float double -> integer, s
after conversion
s = 1Dh  29 decimal

we shift 1 by s times on the left

``  i = 1 << s   thus, i = 1 * 2^29  = 20000000h ( 536'870'912 decimal)``

3 additional checks are needed here. to avoid
overflow after 63 shifts; in the case s > 63 and
checking wether s push i out of the float double capacity.

we reconvert i to float for later use
`` 20000000h integer = 41C0000000000000h float double``

we convert the float decimals we stored above in base2
to float decimals in baseE. this is because we want to use
Taylor series from the e^n. i choosed 10 steps max, and having
-1.0 < decimals < +1.0 works enough good.
NOTE: you can extend the range of action of the
Taylor series up to +-8, stepping it ~20 or more times.

recall Taylor now on e^x :
``  e^x = 1 + x + x^2 / 2! + x^3 / 3! ....  + x^n / n!``

now, according fundamentals of logarithms
``  a) if e^x = 2^q  b) and generally, lgBASE(x)^n = n * lgBASE(x)     then we can extend a) this way,      x * ln(e) = q * ln(2)  c) and thus x = q * ln(2) should verify the a) as true identity.  now, because ln(2) = lg10(2) / lg10(e)    lg10(2) = 0.30102999566398119521373889472449f    lg10(e) = 0.43429448190325179004808384911378f    ln(2)   = 0.69314718055994536943283387715543f  we apply c)    x =  q * ln(2)  where q = r our remainder    x = 0.3 base2 * 0.69314718055994536943283387715543f    x = 3FC9700ADD042628 baseE``

we give this x to the Taylor expansion routine.
to get back
``  3FF3848660139D52 as result of exp() on the remainder 0.3``

finally for the fundamental of powers above
we multiply
`` 41C0000000000000h * 3FF3848660139D52h    n^20 base2      *    n^0.3 base2     = n^20.3 base2  to get back  41C3848660139D52h that corresponds exactly  to our decimal 654904512.1532385``

The resulting assmbly code can be found on my website at

it is ~40 lines of code (my Taylors's exp() code + main routine)
it accepts only +numbers for now, and makes no exaustive check
on the floats. for those and negative values i leave it to the
reader's creativity ( being e^-x essentially 1 / e^x ).
this is the fastest method i know, it should time ~80 cycles totally,
i didnt check it yet. it's not so important.

of some relevance to me was

1) avoid the Intel Approx Math library license
2) avoid things like the cmath library
3) avoid the 2 FPU slow instructions
FYL2X to compute y * log2(x)
F2XM1 to compute 2^x - 1
because 250/300 cycles for 2 instro
it's the insanity, 100%, pure especially
on tests i am doing from huge RND-data outputs.

4) using Taylor series the right way,
because we would need lot of steps to get
the right values on plugging in large x, as in the example e^20.3

but the true-truth is that i am not yet ready
for Chebyshev; simply because i need some time
to understand something more of his genial calculus.
if you have simplified references about him,