Page 276 - Numerical Analysis Using MATLAB and Excel

Solutions to End−of−Chapter Exercises

b. fx() = sin , x f0() = , 0 f' x() = cos , x f' 0() = , 1 f'' x() = – sin , x f'' 0() = , 0
f''' x() = – cos x , f''' 0() = – 1 , and so on. Therefore,

----- –
----- +
f x() = x – x 3 x 5 x 7 …
----- +
n
5!
7!
3!
MATLAB displays the same result.
x=sym('x'); fn=taylor(sin(x)); pretty(fn)
3 5
x - 1/6 x + 1/120 x
c. fx() = sin hx , f0() = , 0 f' x() = cos hx , f' 0() = 1 , f'' x() = sin hx , f'' 0() = , 0
f''' x() = cos hx , f''' 0() = 1 , and so on. Therefore,
3 5 7
f x() = x + x x x …
----- +
----- +
----- +
n
3!
5!
7!
MATLAB displays the same result.
x=sym('x'); fn=taylor(sinh(x)); pretty(fn)
3 5
x + 1/6 x + 1/120 x
8.
f'' a() 2 f''' a() 3
(
)
f x() = fa() + f' a() x – a + ------------ x –( 2! a ) + ------------- x –( 3! a ) + …
n
⁄
)
(
⁄
(
)
a. fx() = 1x , fa() = f – 1 = – , 1 f' x() = – 1 x ⁄ 2 , f' a() = f' – 1 = – , 1 f'' x() = 2x 3 ,
(
(
)
)
f'' a() = f'' – 1 = – 2 , f''' x() = – 6 x ⁄ 4 , f''' a() = f''' – 1 = – 6 , and so on. Therefore,
(
)
–
f x() = – 1 x + 1 – ( x + 1 ) 2 – ( x + 1 ) 3 + …
n
or
f x() = – 2 x ( – – x + 1 ) 2 – ( x + 1 ) 3 + …
n
MATLAB displays the same result.
x=sym('x'); y=1/x; z=taylor(y,4,−1); pretty(z)
2 3
-2 - x - (x + 1) - (x + 1)
(
⁄
)
⁄
)
⁄
(
⁄
b. fx() = sin , x fa() = f – π 4 = – 22 , f' x() = cos , x f' a() = f' – π 4 = 22 ,
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−59
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