Taylor Series Problems
APPM 2350
Spring 2014
1. Use Taylor’s formula for f (x, y) to find second order and third order approximations
of f near the origin.
(a) f (x, y) = x3 y 3 + sin x cos y
Solution:
fx =3x2 y 3 + cos x cos y,
fy =3x3 y 2 − sin x sin y.
fxx =6xy 3 − sin x cos y,
fxy =9x2 y 2 − cos x sin y,
fyy =6x3 y − sin x cos y.
fxxx =6y 3 − cos x cos y,
fxxy =18xy 2 + sin x sin y,
fxyy =18x2 y − cos x cos y,
fyyy =6x3 + sin x sin y.
T2 (x, y) =0 + (1)x + (0)y +
1
(0)x2 + 2(0)xy + (0)y 2
2
=x
1
(−1)x3 + 3(0)x2 y + 3(−1)xy 2 + (0)y 3
6
1
1
= x − x3 − xy 2
6
2
T3 (x, y) =x +
Note: You will get the same answer if you instead find T2 and T3 for the function
g(x, y) = sin x cos y, because x3 y 3 is already a Taylor series of degree 6.
(b) f (x, y) = ln (3x + 2y + 1)
Solution:
3
,
3x + 2y + 1
2
.
fy =
3x + 2y + 1
fx =
−9
,
(3x + 2y + 1)2
−6
,
=
(3x + 2y + 1)2
−4
=
.
(3x + 2y + 1)2
fxx =
fxy
fyy
9 [2 (3x + 2y + 1) 3]
(3x + 2y + 1)4
9 [2 (3x + 2y + 1) 2]
=
(3x + 2y + 1)4
6 [2 (3x + 2y + 1) 2]
=
(3x + 2y + 1)4
4 [2 (3x + 2y + 1) 2]
=
(3x + 2y + 1)4
fxxx =
fxxy
fxyy
fyyy
T2 (x, y) =0 + 3x + 2y +
54
,
(3x + 2y + 1)3
36
=
,
(3x + 2y + 1)3
24
=
,
(3x + 2y + 1)3
16
=
.
(3x + 2y + 1)3
=
1
(−9)x2 + 2(−6)xy + (−4)y 2
2
9
= 3x + 2y − x2 − 6xy − 2y 2
2
9 2
1
2
T3 (x, y) = 3x + 2y − x − 6xy − 2y + (54)x3 + 3(36)x2 y + 3(24)xy 2 + (16)y 3
2
6
9
8
= 3x + 2y − x2 − 6xy − 2y 2 + 9x3 + 18x2 y + 12xy 2 + y 3
2
3
Alternate Solution:
ln [1 + (3x + 2y)] = (3x + 2y) −
(3x + 2y)2 (3x + 2y)3
+
− ···
2
3
9
8
= 3x + 2y − x2 − 6xy − 2y 2 + 9x3 + 18x2 y + 12xy 2 + y 3 − · · ·
2
3
2. Use Taylor’s formula for f (x, y) to find second order and third order approximations
of f near the origin. Estimate the error in using the second order approximation if
|x| ≤ 0.2 and |y| ≤ 0.1.
(a) f (x, y) = ex+y sin(x + y)
Solution:
fx =ex+y cos(x + y) + sin(x + y)ex+y
=ex+y [sin(x + y) + cos(x + y)] ,
fy =fx .
fxx =ex+y [cos(x + y) − sin(x + y)] + [sin(x + y) + cos(x + y)] ex+y
=2ex+y cos(x + y),
fxy =fxx ,
fyy =fxx .
fxxx =2 ex+y [− sin(x + y)] + cos(x + y)ex+y
fxxy
fxyy
fyyy
=2ex+y [cos(x + y) − sin(x + y)] ,
=fxxx ,
=fxxx ,
=fxxx .
1
(2)x2 + 2(2)xy + (2)y 2
2
2
= x + y + x + 2xy + y 2
T2 (x, y) =0 + (1)x + (1)y +
1
T3 (x, y) = x + y + x2 + 2xy + y 2 + (2)x3 + 3(2)x2 y + 3(2)xy 2 + (2)y 3
6
1
1
= x + y + x2 + 2xy + y 2 + x3 + x2 y + xy 2 + y 3
3
3
1
|fxxx | |x|3 + 3 |fxxy | |x|2 |y| + 3 |fxyy | |x||y|2 + |fyyy | |y|3
6
1
≤ 1(.2)3 + 3(.2)2 (.1) + 3(.2)(.1)2 + 1(.1)3 2e0.3 [cos 0 + sin(.3)]
6
≈ 0.0157
|R2 (x, y)| ≤
(b) f (x, y) = cos (x2 + y 2 )
Solution:
fx = − 2x sin x2 + y 2 ,
fy = − 2y sin x2 + y 2 .
fxx = − 2 x 2x cos x2 + y 2 + sin x2 + y 2
= − 2 2x2 cos x2 + y 2 + sin x2 + y 2 ,
fxy = − 4xy cos x2 + y 2 ,
fyy = − 2 y 2y cos x2 + y 2 + sin x2 + y 2
= − 2 2y 2 cos x2 + y 2 + sin x2 + y 2 .
fxxx = − 2 2x2 −2x sin x2 + y 2 + cos x2 + y 2 (4x) + 2x cos x2 + y 2
= − 4x −2x2 sin x2 + y 2 + 3 cos x2 + y 2 ,
fxxy = − 2 −4x2 y sin x2 + y 2 + 2y cos x2 + y 2
= − 4y −2x2 sin x2 + y 2 + cos x2 + y 2 ,
fxyy = − 4x y −2y sin x2 + y 2 + cos x2 + y 2
= − 4x −2y 2 sin x2 + y 2 + cos x2 + y 2 ,
fyyy = − 2 2y 2 −2y sin x2 + y 2 + cos x2 + y 2 (4y) + 2y cos x2 + y 2
= − 4y −2y 2 sin (x + y) + 3 cos x2 + y 2 .
All of these derivatives are zero at the origin, so T2 (x, y) = 1 and T3 (x, y) = 1
1
|fxxx | |x|3 + 3 |fxxy | |x|2 |y| + 3 |fxyy | |x||y|2 + |fyyy | |y|3
6
1
≤ (4)(.2) 2(.2)2 sin (.2)2 + (.1)2 + 3 (.2)3
6
+3(4)(.1) 2(.2)2 sin (.2)2 + (.1)2 + 1 (.2)2 (.1)
+3(4)(.2) 2(.1)2 sin (.2)2 + (.1)2 + 1 (.2)(.1)2
+(4)(.1) 2(.1)2 sin (.2)2 + (.1)2 + 3 (.1)3
|R2 (x, y)| ≤
≈ 0.005
Note: Since T2 = T3 , the error is probably much lower than this.