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i.e.themostimportantcomponent:
1
ydx
dy
±
ln
(
|
k
1
1
–
+
x
x
N
|
)
–
1
–
2
x
x
2
ofthesought-forfunctionT(x):
Tx
()
=
xdy
ydx
±
x
ln
(
|
k
1
1
–
+
x
x
N
|
)
–
1
2
–
x
x
2
2
ThecorrectnessoftheaboveexpressionhasbeenverifiedbymeansofthefollowingMATLABscript:
%RELATIVEERRORPROPAGATIONTHROUGH
%FORMULAy=((1-x)/(1+x))^x
symsx
y=((1-x)/(1+x))^x;
T=x*diff(y,x)/y;
fplot(log10(abs(T)),[0,1]);gridon
xlabel('x');ylabel('log[|T(x)|]')
title('y=[(1-x)/(1+x)]^x')
T1=x*log((1-x)/(1+x))-2*x^2/(1-x^2);
fprintf('T1(x)-T(x)=%s\n',simplify(T1-T))
%referenceexpressionforT(x)
%testedexpressionforT(x)
T1(x)-T(x)=0
Problem1.7:DeterminethefunctionT(x),characterisingthepropagationofthedataerrorthroughthefol-
lowingformula:
y
±f
L
tan
()
x
1
Jfor
x
x
E
(
0,
π
2
)
Solution#1:TheproblemmaybeconsiderablysimplifiedbycomputingthenaturallogarithmoftheLHS
andRHSoftheanalysedformula:
ln
()
y
±
x
lntan
f
L
()
x
1
J
sincethedifferentiationofbothsidesofthetransformedfunctionyields:
1
ydx
dy
±
lntan
f
L
()
x
1+
J
x
tan
1
()
x
cos
2
1
()
x
±
lntan
f
L
()
x
1+
J
sin2
2
()
x
x
i.e.themostimportantcomponentofthesought-forfunctionT(x):
Tx
()
=
xdy
ydx
±
x
lntan
f
L
()
x
1+
J
sin2
2
()
x
2
x
Solution#2:Alternatively,thefunctionT(x)maybedeterminedinthefollowingway:
Tx
()
=
d
d
(
(
ln
ln
()
()
y
x
)
)
±
du
dv
where
u
=
ln
()
x
and
v
=
ln
()
y
±
e
u
lntan
(
()
e
u
)
.
Hence:
Tx
()
±
du
dv
±
...
±
e
u
lntan
(
()
e
u
)
+
e
u
tan
1
()
e
u
cos
2
1
()
e
u
e
u
±
x
lntan
f
L
()
x
1+
J
sin2
2
()
x
2
x
18
