7/28/2019 Julia Working Note: Zero to Zero
1/2
Department of Chemistry, Van Voorhis groupJiahao Chen
Research notes:April 17, 2013, page 1 of 2
Massachusetts Institute of Technology77 Massachusetts Avenue,
Room 6-228Cambridge, Massachusetts 02139-4307
Email [email protected]
Working out 00
Kahan case 1
Step 1. Expand the exponential so that
zw = limz
exp (w ln )
z0.0 = limw0.0
limz
exp (w ln )
= 1+
lim
w0.0w
limz
ln
+O
w2
Step 2. Now we want to take the limit z 0. Rewrite = rei and
interpret z 0 to mean r 0+ withthe phase as yet undefined. Then we
have:
0.00.0 = 1+
lim
w0.0w
limr0+
ln rei+O
w2
= 1+ i
lim
w0.0w
limr0+
ln r
+O
w2
Step 3. Since the limits ofw and r are now independent, we can
safely take the w limit first to see thatthe second term must go to
zero. However, we can refine this statement, since for 0 < r
< 1 we haveln r < 0. Furthermore, if we write out w = x+ iy,
then we get
0.00.0 = 1 +
lim
y0.0y
(something negative) + i
lim
w0.0x
(something negative)
Thus the answer should be
0.00.0 = limw0
limz0
zw = 1 isgn arg (w) sgn Re (z) lim0+
where is the phase of the complex zero in the base and is an
arbitrary positive number put in sothat the imaginary part goes to
zero.
Kahan case 2
Step 1. Take z and w to be implicit functions of, then near we
can write
z () = z () + () z
() +O()2
w () = w () + ()w () +O
()2
April 17, 2013 Working out 00 Page 1 of 2
7/28/2019 Julia Working Note: Zero to Zero
2/2
Department of Chemistry, Van Voorhis groupJiahao Chen
Research notes:April 17, 2013, page 2 of 2
Massachusetts Institute of Technology77 Massachusetts Avenue,
Room 6-228Cambridge, Massachusetts 02139-4307
Email [email protected]
Then expand zw in powers of():
zw = limz
exp (w ln ) = 1 + w limz
ln + = 1 +w () + ()w ()
lnz () + () z ()
+ . . .
= 1 +w () + ()w ()
lnz ()
1 + () z
()z ()
+ . . .
+ . . .
= 1 +w () + ()w ()
lnz () + ln
1 + ()
z ()
z ()+ . . .
+ . . .
= 1 +w () + ()w ()
lnz () + ()
z ()
z ()+ . . .
+ . . .
= 1 + w () lnz () + ()
w () lnz () + w ()
z ()
z ()
+ . . .
First take the limit , so that the term in () goes away. Then we
are left with
lim
z ()w() = 1 +w () lnz ()
We have reached exactly the same point as Step 2 of Case 1. The
only difference is that the limit z is worked out trivially.
April 17, 2013 Working out 00 Page 2 of 2
