Hyperbolic functions 49
8. 2 sh x + 3 ch x = 5
[0.6389 or −2.2484]
9. 4 th x − 1 = 0
[ 0 .2554]
10. A chain hangs so that its shape is of the
form y = 56 cosh
x
56
. Determine, correct to
4 significant figures, (a) the value of y when
x is 35, and (b) the value of x when y is 62.35
[(a) 67.30 (b) ±26.42]
5.5 Series expansions for cosh x and
sinh x
By definition,
e
x
= 1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+ · · ·
from Chapter 4.
Replacing x by −x gives:
e
−x
= 1 − x +
x 2
2!
−
x 3
3!
+
x 4
4!
−
x 5
5!
+ · · · .
cosh x =
1
2
(e
x
+ e
−x
)
=
1
2
1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+ · · ·
+
1 − x +
x
2
2!
−
x
3
3!
+
x
4
4!
−
x
5
5!
+ · · ·
=
1
2
2 +
2x 2
2!
+
2x 4
4!
+ · · ·
i.e. cosh x = 1 +
x 2
2!
+
x 4
4!
+· · · (which is valid for all
values of x). cosh x is an even function and contains
only even powers of x in its expansion.
sinh x =
1
2
(e
x
− e
−x
)
=
1
2
1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+· · ·
−
1 − x +
x 2
2!
−
x 3
3!
+
x 4
4!
−
x 5
5!
+· · ·
=
1
2
2x +
2x 3
3!
+
2x 5
5!
+ · · ·
i.e. sinh x = x +
x 3
3!
+
x 5
5!
+· · · (which is valid for all
values of x). sinh x is an odd function and contains only
odd powers of x in its series expansion.
Problem 23. Using the series expansion for ch x
evaluate ch 1 correct to 4 decimal places.
ch x = 1 +
x 2
2!
+
x 4
4!
+ · · ·from above
Let x = 1,
then ch 1 = 1 +
1 2
2 × 1
+
1 4
4 ×3 × 2 × 1
+
1
6
6 ×5 × 4 × 3 ×2 × 1
+ · · ·
= 1 + 0.5 + 0.04167 + 0.001389 + · · ·
i.e. ch 1 = 1.5431, correct to 4 decimal places,
which may be checked by using a calculator.
Problem 24. Determine, correct to 3 decimal
places, the value of sh 3 using the series expansion
for sh x.
sh x = x +
x 3
3!
+
x 5
5!
+ · · · from above
Let x = 3, then
sh 3 = 3 +
3 3
3!
+
3 5
5!
+
3 7
7!
+
3 9
9!
+
3 11
11!
+ · · ·
= 3 + 4.5 + 2.025 + 0.43393 + 0.05424
+ 0.00444 + · · ·
i.e. sh 3 = 10.018, correct to 3 decimal places.
Problem 25. Determine the power series for
2 ch
θ
2
− sh 2θ as far as the term in θ 5 .
In the series expansion for ch x, let x =
θ
2
then:
2 ch
θ
2
= 2
1 +
(θ/2)
2
2!
+
(θ/2)
4
4!
+ · · ·
= 2 +
θ 2
4
+
θ 4
192
+ · · ·
8. 2 sh x + 3 ch x = 5
[0.6389 or −2.2484]
9. 4 th x − 1 = 0
[ 0 .2554]
10. A chain hangs so that its shape is of the
form y = 56 cosh
x
56
. Determine, correct to
4 significant figures, (a) the value of y when
x is 35, and (b) the value of x when y is 62.35
[(a) 67.30 (b) ±26.42]
5.5 Series expansions for cosh x and
sinh x
By definition,
e
x
= 1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+ · · ·
from Chapter 4.
Replacing x by −x gives:
e
−x
= 1 − x +
x 2
2!
−
x 3
3!
+
x 4
4!
−
x 5
5!
+ · · · .
cosh x =
1
2
(e
x
+ e
−x
)
=
1
2
1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+ · · ·
+
1 − x +
x
2
2!
−
x
3
3!
+
x
4
4!
−
x
5
5!
+ · · ·
=
1
2
2 +
2x 2
2!
+
2x 4
4!
+ · · ·
i.e. cosh x = 1 +
x 2
2!
+
x 4
4!
+· · · (which is valid for all
values of x). cosh x is an even function and contains
only even powers of x in its expansion.
sinh x =
1
2
(e
x
− e
−x
)
=
1
2
1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+· · ·
−
1 − x +
x 2
2!
−
x 3
3!
+
x 4
4!
−
x 5
5!
+· · ·
=
1
2
2x +
2x 3
3!
+
2x 5
5!
+ · · ·
i.e. sinh x = x +
x 3
3!
+
x 5
5!
+· · · (which is valid for all
values of x). sinh x is an odd function and contains only
odd powers of x in its series expansion.
Problem 23. Using the series expansion for ch x
evaluate ch 1 correct to 4 decimal places.
ch x = 1 +
x 2
2!
+
x 4
4!
+ · · ·from above
Let x = 1,
then ch 1 = 1 +
1 2
2 × 1
+
1 4
4 ×3 × 2 × 1
+
1
6
6 ×5 × 4 × 3 ×2 × 1
+ · · ·
= 1 + 0.5 + 0.04167 + 0.001389 + · · ·
i.e. ch 1 = 1.5431, correct to 4 decimal places,
which may be checked by using a calculator.
Problem 24. Determine, correct to 3 decimal
places, the value of sh 3 using the series expansion
for sh x.
sh x = x +
x 3
3!
+
x 5
5!
+ · · · from above
Let x = 3, then
sh 3 = 3 +
3 3
3!
+
3 5
5!
+
3 7
7!
+
3 9
9!
+
3 11
11!
+ · · ·
= 3 + 4.5 + 2.025 + 0.43393 + 0.05424
+ 0.00444 + · · ·
i.e. sh 3 = 10.018, correct to 3 decimal places.
Problem 25. Determine the power series for
2 ch
θ
2
− sh 2θ as far as the term in θ 5 .
In the series expansion for ch x, let x =
θ
2
then:
2 ch
θ
2
= 2
1 +
(θ/2)
2
2!
+
(θ/2)
4
4!
+ · · ·
= 2 +
θ 2
4
+
θ 4
192
+ · · ·