58 Hyperbolic
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Hyperbolic Functions (Euler) ex + e−x 2 e − e−x
will be called the Hyperbolic Cosine, cosh x.
x
will be called the Hyperbolic Sine,
2 Symmetrically, sinh and cosh are derivatives,
sinh x.
and also antiderivatives, of each other. The Hyperbolic Tangent is their quotient: ex − e−x sinh x = x . tanh x ≡ −x cosh x e +e We also have 1 1 cosh x sech x ≡ , cosech x ≡ , cotanh x ≡ . cosh x sinh x sinh x
cosh2x − sinh2x = 1. ex + e−x 2 ex − e−x 2 Proof. − 2 2 ex + e−x x −x x e −e ex − e−x e + e−x + − = 2 2 2 2 2 ex 2 e−x = 1. = 2 2 d dx
Proof.
d sinh x dx cosh x
tanh x = sech 2x. =
(sinh x)0 cosh x − (cosh x)0 sinh x
cosh2x 1 cosh2x − sinh2x = = 2 cosh x cosh2x
y
1 (cos t, sin t)
x2 + y 2 = 1 Area =
0 0
t 2
1
x
(cosh t, sinh t)
y
1
x2 − y 2 = 1 Area =
0 0
t 2
1
x
x
1
x2 + y 2 = 1
Area =
x= 0 0
y = sin u dy = cos u du p 1 − y 2 = cos u
t 2
cos t sin t
(cos t, sin t)
y 1
y
Z =
=
= =
Z 1 2 1 2 t 2
sin t y=0 t
p 1 − y 2 dy
cos u cos u du u=0
Z
− −
t u=0
1 + cos 2u du − +
t +
sin 2t 2
2 sin t cos t 4
− −
Z
sin t y=0
cos t sin t
cos t
sin t Z sin t
cos t sin t
y dy
y dy
y=0
"
y
2
2
#sin t
y=0
cos t sin2t sin t
2
cos t sin t 2
=
t 2
(cosh t, sinh t) x2 − y 2 = 1 x
1
y = sinh u
1
Area =
t 2
x=
0 0
dy = cosh u du p 1 + y 2 = cosh u cosh t sinh t
y
0
0
1
1
y
Z =
=
= =
Z Z 1 2 t 2
sinh t y=0 t
p 1 + y 2 dy 2
cosh u du
−
u=0 t
1 + cosh 2u 2
u=0
t +
+
du −
sinh 2t 2
2 sinh t cosh t 4
−
− −
Z
sinh t y=0
cosh t sinh t
cosh t sinh t Z sinh t
cosh t sinh t
y dy
y dy
y=0
"
y2 2
#sinh t y=0
cosh t sinh2t sinh t
2
cosh t sinh t 2
=
t 2
sinh (s + t) = sinh s cosh t + cosh s sinh t es − e−s et + e−t
Proof. =
2 2 2 eset − e−set + ese−t − e−se−t +
= =
+
es + e−s et − e−t 2
4 eset + e−set − ese−t − e−se−t 4
2 eset − 2 e−se−t 4 e(s+t) − e−(s+t) 2
sinh 2s = 2 sinh s cosh s
cosh (s + t) = cosh s cosh t + sinh s sinh t es + e−s et + e−t
Proof. =
2 2 2 eset + e−set + ese−t + e−se−t +
=
+
es − e−s et − e−t 2
4 eset − e−set − ese−t + e−se−t 4
2 eset + 2 e−se−t 4
=
e(s+t) + e−(s+t) 2
cosh 2s = cosh2s + sinh2s = 2 cosh2s − 1, = 1 + 2 sinh2s,
using using
sinh2s = cosh2s − 1,
cosh2s = 1 + sinh2s.
x
e = 1 + x + e
−x
ex + e−x 2
= 1 − x +
+
cos x = 1
2
2 x2 2
+ −
x3 6 x3 6
+ +
x4 24 x4 24
+ −
x5 120 x5 120
+ ... + ...
= =
∞ X xn
n=0 ∞ X n=0
n!
(−1)nxn n!
=
cosh x = 1
ex − e−x
x2
−
x2
+
2 x2
+
2
x4
+ ...
24 x4
+ ...
24
= =
∞ X x2n
n=0 ∞ X n=0
(2n)! (−1)nx2n (2n)!
=
sinh x = sin x =
x x
+ −
x3 6 x3 6
+ +
x5 120 x5 120
+ ... + ...
= =
∞ X
n=0 ∞ X n=0
x2n+1 (2n + 1)! (−1)nx2n+1 (2n + 1)!
Recall: i2 = −1, x
e = 1 + x e e
ix
ix
= 1 + ix = 1
i3 = −i, + − −
x2 2 x2 2
cos ix = 1
− +
−
x3
+
6 ix3 6
x2
x2
+ +
2
+ i x cos x = 1
+
i4 = 1,
−
x3 +
2 x2
+
sin x =
x
−
sin ix =
ix
+
x
3
6 ix3
We also have cosh ix = cos x, tan ix = i tanh x,
6
x4 24 x4 24
+ +
etc.
x5 120 ix5 120
x4
+ +
x4
x5 120
24 x4 24 + +
x5 120 ix5 120
=
...
=
...
∞ X xn
n!
n=0 ∞ n n X
i x
n=0
+
24 +
6
2
i5 = i,
...
+
...
+
...
+
...
+
...
+
...
sinh ix = i sin x, tanh ix = i tan x.
n!
= cos x + i sin x
= cosh x
= i sinh x
ex = cosh x + sinh x eix = cos x + i sin x e−ix = cos x − i sin x cos x = sin x =
eix + e−ix 2 eix − e−ix 2i
eiπ = cos π + i sin π = −1
eiπ + 1 = 0
If y equals arcsinh x, then x equals sinh y =
ey − e−y 2
.
2x = ey − e−y
ey − 2x − e−y = 0
e2y − 2xey − 1 = 0
√ 2x ± 4x2 + 4
p = x + x2 + 1 ey = 2p 2 y = ln x + x + 1 √ 2 arcsinh x = ln x + x + 1 1
2x
√
x2 + 1
x
+√ 1+ √ √ 2 2 1 2 x +1 x +1 x2 + 1 0 = =√ (arcsinh x) = √ √ 2 2 x+ x +1 x+ x +1 1 + x2
If y equals arccosh x, then x equals cosh y =
ey + e−y 2
.
2x = ey + e−y ey − 2x + e−y = 0
e2y − 2xey + 1 = 0
2x ±
√
4x2 − 4
p ey = = x + x2 − 1 2p y = ln x + x2 − 1 √ arccosh x = ln x + x2 − 1 for x ≥ 1 1
2x
√
x2 − 1
x
1+ √ +√ √ 2 2 1 2 x −1 x −1 x2 − 1 0 = =√ (arccosh x) = √ √ x + x2 − 1 x + x2 − 1 x2 − 1
ey − e−y
If y equals arctanh x, then x equals tanh y = y . −y e +e y −y y −y x=e −e e +e xey + xe−y = ey − e−y
0 = (1 − x)ey − (1 + x)e−y
(1 + x)e−y = (1 − x)ey 1+x = e2y 1−x 1+x 2y = ln 1− x 1 1+x y = ln 2 1−x 1+x 1 arctanh x = ln for −1 < x < 1 2 1−x
1
0
(arctanh x) =
2
ln
1+x 1−x
!0
1
ln (1 + x) − ln (1 − x) 2 1 −1 1 − = 2 1+x 1−x =
= = (arctanh x)0 =
1 (1 − x) + (1 + x) 2 1
(1 + x)(1 − x) 2
2 (1 + x)(1 − x) 1 1 − x2
0
Integral Substitutions using
a2 cosh2t − a2 sinh2t = a2, a2 tanh2t + a2 sech2t = a2
If x = a sinh t : p p x2 + a2 = a2 sinh2t + a2 p = a sinh2t + 1 = a cosh t t = arcsinh
x a
= ln
x a
+
√
x2
+
a
a2
p = ln (x + x2 + a2) − ln a
If x = a cosh t : p p x2 − a2 = a2 cosh2t − a2 p = a cosh2t − 1 = a sinh t t = arccosh
x a
= ln
x a
+
= ln (x +
√
p
x2
−
a
a2
x2 − a2) − ln a
If x = a tanh t : p p a 2 − x2 = a2 − a2 tanh2t p = a 1 − tanh2t = a sech t a+x 1 x = ln t = arctanh a 2 a−x
Every nonzero complex number x + iy can be written as its p absolute value r = x2 + y 2, which is real, times a number with absolute value equal to 1, which lies on the unit circle and which can be written as cos θ + i sin θ, where θ ± 2kπi is a x x certain angle satisfying cos θ = and sin θ = . Call θ the r r argument of z : θ = arg (z). Let ρ equal ln r. Since we have z = x + iy = r(cos θ + i sin θ) = eρeiθ = eρ+iθ , we can define the complex logarithm as a (multi- or singlevalued) inverse of the complex exponential as follows: ln z = ln x + iy = ρ + iθ =
1 2
ln (x2 + y 2) + i arg (z)
With this complex logarithm, the inverses of the (circular) trigonometric and hyperbolic functions can be listed:
arcsin z = arctan z =
arcsec z =
1 i
ln z +
1 2i 1 i
ln
ln
p
1−
1 + iz
z2
,
1 − iz √ 1 + 1 − z2 z
p 2 arc sinh z = ln z + z + 1 , arctanhz =
1
2
ln
arcsechz = ln
,
1+z
,
1−z √ 1 + 1 − z2 z
arccos z = arc cot z =
, arc csc z =
1 i
ln z +
1 2i 1 i
ln
ln
p
z2
z+i
−1 ,
,
z−i √ 1 + z2 − 1 z
p 2 arccoshz = ln z + z − 1 , arccothz =
1
2
ln
, arccosechz = ln
z+1
,
z−1 √ 1 + z2 + 1 z
.
,
will be called the Hyperbolic Cosine, cosh x.
x
will be called the Hyperbolic Sine,
2 Symmetrically, sinh and cosh are derivatives,
sinh x.
and also antiderivatives, of each other. The Hyperbolic Tangent is their quotient: ex − e−x sinh x = x . tanh x ≡ −x cosh x e +e We also have 1 1 cosh x sech x ≡ , cosech x ≡ , cotanh x ≡ . cosh x sinh x sinh x
cosh2x − sinh2x = 1. ex + e−x 2 ex − e−x 2 Proof. − 2 2 ex + e−x x −x x e −e ex − e−x e + e−x + − = 2 2 2 2 2 ex 2 e−x = 1. = 2 2 d dx
Proof.
d sinh x dx cosh x
tanh x = sech 2x. =
(sinh x)0 cosh x − (cosh x)0 sinh x
cosh2x 1 cosh2x − sinh2x = = 2 cosh x cosh2x
y
1 (cos t, sin t)
x2 + y 2 = 1 Area =
0 0
t 2
1
x
(cosh t, sinh t)
y
1
x2 − y 2 = 1 Area =
0 0
t 2
1
x
x
1
x2 + y 2 = 1
Area =
x= 0 0
y = sin u dy = cos u du p 1 − y 2 = cos u
t 2
cos t sin t
(cos t, sin t)
y 1
y
Z =
=
= =
Z 1 2 1 2 t 2
sin t y=0 t
p 1 − y 2 dy
cos u cos u du u=0
Z
− −
t u=0
1 + cos 2u du − +
t +
sin 2t 2
2 sin t cos t 4
− −
Z
sin t y=0
cos t sin t
cos t
sin t Z sin t
cos t sin t
y dy
y dy
y=0
"
y
2
2
#sin t
y=0
cos t sin2t sin t
2
cos t sin t 2
=
t 2
(cosh t, sinh t) x2 − y 2 = 1 x
1
y = sinh u
1
Area =
t 2
x=
0 0
dy = cosh u du p 1 + y 2 = cosh u cosh t sinh t
y
0
0
1
1
y
Z =
=
= =
Z Z 1 2 t 2
sinh t y=0 t
p 1 + y 2 dy 2
cosh u du
−
u=0 t
1 + cosh 2u 2
u=0
t +
+
du −
sinh 2t 2
2 sinh t cosh t 4
−
− −
Z
sinh t y=0
cosh t sinh t
cosh t sinh t Z sinh t
cosh t sinh t
y dy
y dy
y=0
"
y2 2
#sinh t y=0
cosh t sinh2t sinh t
2
cosh t sinh t 2
=
t 2
sinh (s + t) = sinh s cosh t + cosh s sinh t es − e−s et + e−t
Proof. =
2 2 2 eset − e−set + ese−t − e−se−t +
= =
+
es + e−s et − e−t 2
4 eset + e−set − ese−t − e−se−t 4
2 eset − 2 e−se−t 4 e(s+t) − e−(s+t) 2
sinh 2s = 2 sinh s cosh s
cosh (s + t) = cosh s cosh t + sinh s sinh t es + e−s et + e−t
Proof. =
2 2 2 eset + e−set + ese−t + e−se−t +
=
+
es − e−s et − e−t 2
4 eset − e−set − ese−t + e−se−t 4
2 eset + 2 e−se−t 4
=
e(s+t) + e−(s+t) 2
cosh 2s = cosh2s + sinh2s = 2 cosh2s − 1, = 1 + 2 sinh2s,
using using
sinh2s = cosh2s − 1,
cosh2s = 1 + sinh2s.
x
e = 1 + x + e
−x
ex + e−x 2
= 1 − x +
+
cos x = 1
2
2 x2 2
+ −
x3 6 x3 6
+ +
x4 24 x4 24
+ −
x5 120 x5 120
+ ... + ...
= =
∞ X xn
n=0 ∞ X n=0
n!
(−1)nxn n!
=
cosh x = 1
ex − e−x
x2
−
x2
+
2 x2
+
2
x4
+ ...
24 x4
+ ...
24
= =
∞ X x2n
n=0 ∞ X n=0
(2n)! (−1)nx2n (2n)!
=
sinh x = sin x =
x x
+ −
x3 6 x3 6
+ +
x5 120 x5 120
+ ... + ...
= =
∞ X
n=0 ∞ X n=0
x2n+1 (2n + 1)! (−1)nx2n+1 (2n + 1)!
Recall: i2 = −1, x
e = 1 + x e e
ix
ix
= 1 + ix = 1
i3 = −i, + − −
x2 2 x2 2
cos ix = 1
− +
−
x3
+
6 ix3 6
x2
x2
+ +
2
+ i x cos x = 1
+
i4 = 1,
−
x3 +
2 x2
+
sin x =
x
−
sin ix =
ix
+
x
3
6 ix3
We also have cosh ix = cos x, tan ix = i tanh x,
6
x4 24 x4 24
+ +
etc.
x5 120 ix5 120
x4
+ +
x4
x5 120
24 x4 24 + +
x5 120 ix5 120
=
...
=
...
∞ X xn
n!
n=0 ∞ n n X
i x
n=0
+
24 +
6
2
i5 = i,
...
+
...
+
...
+
...
+
...
+
...
sinh ix = i sin x, tanh ix = i tan x.
n!
= cos x + i sin x
= cosh x
= i sinh x
ex = cosh x + sinh x eix = cos x + i sin x e−ix = cos x − i sin x cos x = sin x =
eix + e−ix 2 eix − e−ix 2i
eiπ = cos π + i sin π = −1
eiπ + 1 = 0
If y equals arcsinh x, then x equals sinh y =
ey − e−y 2
.
2x = ey − e−y
ey − 2x − e−y = 0
e2y − 2xey − 1 = 0
√ 2x ± 4x2 + 4
p = x + x2 + 1 ey = 2p 2 y = ln x + x + 1 √ 2 arcsinh x = ln x + x + 1 1
2x
√
x2 + 1
x
+√ 1+ √ √ 2 2 1 2 x +1 x +1 x2 + 1 0 = =√ (arcsinh x) = √ √ 2 2 x+ x +1 x+ x +1 1 + x2
If y equals arccosh x, then x equals cosh y =
ey + e−y 2
.
2x = ey + e−y ey − 2x + e−y = 0
e2y − 2xey + 1 = 0
2x ±
√
4x2 − 4
p ey = = x + x2 − 1 2p y = ln x + x2 − 1 √ arccosh x = ln x + x2 − 1 for x ≥ 1 1
2x
√
x2 − 1
x
1+ √ +√ √ 2 2 1 2 x −1 x −1 x2 − 1 0 = =√ (arccosh x) = √ √ x + x2 − 1 x + x2 − 1 x2 − 1
ey − e−y
If y equals arctanh x, then x equals tanh y = y . −y e +e y −y y −y x=e −e e +e xey + xe−y = ey − e−y
0 = (1 − x)ey − (1 + x)e−y
(1 + x)e−y = (1 − x)ey 1+x = e2y 1−x 1+x 2y = ln 1− x 1 1+x y = ln 2 1−x 1+x 1 arctanh x = ln for −1 < x < 1 2 1−x
1
0
(arctanh x) =
2
ln
1+x 1−x
!0
1
ln (1 + x) − ln (1 − x) 2 1 −1 1 − = 2 1+x 1−x =
= = (arctanh x)0 =
1 (1 − x) + (1 + x) 2 1
(1 + x)(1 − x) 2
2 (1 + x)(1 − x) 1 1 − x2
0
Integral Substitutions using
a2 cosh2t − a2 sinh2t = a2, a2 tanh2t + a2 sech2t = a2
If x = a sinh t : p p x2 + a2 = a2 sinh2t + a2 p = a sinh2t + 1 = a cosh t t = arcsinh
x a
= ln
x a
+
√
x2
+
a
a2
p = ln (x + x2 + a2) − ln a
If x = a cosh t : p p x2 − a2 = a2 cosh2t − a2 p = a cosh2t − 1 = a sinh t t = arccosh
x a
= ln
x a
+
= ln (x +
√
p
x2
−
a
a2
x2 − a2) − ln a
If x = a tanh t : p p a 2 − x2 = a2 − a2 tanh2t p = a 1 − tanh2t = a sech t a+x 1 x = ln t = arctanh a 2 a−x
Every nonzero complex number x + iy can be written as its p absolute value r = x2 + y 2, which is real, times a number with absolute value equal to 1, which lies on the unit circle and which can be written as cos θ + i sin θ, where θ ± 2kπi is a x x certain angle satisfying cos θ = and sin θ = . Call θ the r r argument of z : θ = arg (z). Let ρ equal ln r. Since we have z = x + iy = r(cos θ + i sin θ) = eρeiθ = eρ+iθ , we can define the complex logarithm as a (multi- or singlevalued) inverse of the complex exponential as follows: ln z = ln x + iy = ρ + iθ =
1 2
ln (x2 + y 2) + i arg (z)
With this complex logarithm, the inverses of the (circular) trigonometric and hyperbolic functions can be listed:
arcsin z = arctan z =
arcsec z =
1 i
ln z +
1 2i 1 i
ln
ln
p
1−
1 + iz
z2
,
1 − iz √ 1 + 1 − z2 z
p 2 arc sinh z = ln z + z + 1 , arctanhz =
1
2
ln
arcsechz = ln
,
1+z
,
1−z √ 1 + 1 − z2 z
arccos z = arc cot z =
, arc csc z =
1 i
ln z +
1 2i 1 i
ln
ln
p
z2
z+i
−1 ,
,
z−i √ 1 + z2 − 1 z
p 2 arccoshz = ln z + z − 1 , arccothz =
1
2
ln
, arccosechz = ln
z+1
,
z−1 √ 1 + z2 + 1 z
.
,
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