Phgsics 325, Fall 2006
Solutions to Homework Assignment
#I
i)
Taylor expansion of erf (r) :
#
f;
e"u'du. we need the first three nonzero
terms,
evaiuatedatr:0.
erf(0)
: g
erf'

rtu:"
by the fundamental
theorem of calculus.
erf,
:
h
u, r
:
0.
erf't =
rtlZr""]
which
vanishes
at c:0,
er.ft"
:
hlr"n
+ 4rr","1which
hasthe
value
ff
at r
:
0.
erfrrrt

ftl+re,'
*8xe,2

Brt",'l which
,runi.n",
at r
=
0.
:: {""' ;,k
[tle"
+ 4e" + Be,'' I6r2e,'

24rze,'
*
r6rae,rl which has
r]re
value
#
at r:0.
The firit three nonzero
terms of the Taylor expansion
are
e
r
f
(
r
)
= o *
h * o  # + o + m
Hete's an elegant alternative solution (easier,
tool).
you
can express
eu,
:
Ino
g#f
,
and integrate
that ,:l.,T,l{,t.rm
to get a series expansion
for eri(x).
erf(x):rtDy)ffi
Reading
off the first'three
terms gives
e r f ( r ) = * 1 "  4 + t l l
Which agrbes' with it
.
u"U'ou.
computation.
Inserting r:
I to the seriesexpansion, erf(I)
= 0.g6b1, which can be compared with
the value looked up in a table (looking up erf(l) in a table is not required for credit),
erf (I) :0.8427.
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 Fall '09
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