1132 Advanced Calculus
Trinity College Dublin
Instructor: Petronela Radu
Thursday, February 21, 2013
Name:
Solutions to Exam 1
1. Write the fourth order Taylor polynomial for f (x) = x2 + cos(πx) at x = 2.
Solution The fourth order polynomial at x = 2 is given by
T4 (x) = f (2) + f 0 (2)(x − 2) +
f 000 (2)
f (4) (2)
f 00 (2)
(x − 2)2 +
(x − 2)3 +
(x − 2)4 .
2!
3!
4!
We have
f (2) = 4 + cos 2π = 5, f 0 (2) = 4 − π sin 2π = 4, f 00 (2) = 2 − π 2 cos 2π = 2 − π 2 ,
f 000 (2) = π 3 sin 2π, f (4) (2) = π 4 cos 2π = π 4 .
Therefore,
T4 (x) = 5 + 4(x − 2) +
π4
(2 − π 2 )
(x − 2)2 +
(x − 2)4 .
2
24
2. Find the radius and interval of convergence for the power series
∞
X
(−1)n−1
(x − 1)2n−1
(2n
−
1)
n=1
Solution: We use the ratio test, so we compute
n−1
(−1)n+1
(x − 1)2n−1 2n+1 (−1)
2
lim
(x − 1)
·
= |x1 | .
n→∞ (2n + 1)
(2n − 1)
We have convergence of the series for |x − 1| < 1, i.e. on the interval centered at 1 of radius 1. At the
endpoints, we obtain the series with general terms:
∞
X
(−1)3n−1
which is an alternating series with a decreasing sequence for the general
(2n − 1)
n=1
term 1/(2n − 1); by the Alternating Series Test, this series is convergent (only conditionally, since
the absolute value gives us a harmonic series);
∞
X
(−1)n−1
• x=2:
which note that it is exactly the same series as above, hence, conditionally
(2n − 2)
n=1
convergent.
• x=0:
To summarize, the series converges absolutely inside the interval (0, 2) and it converges conditionally
at the endpoints x = 0, 2.
3. Find the partials
∂f
∂f
and
evaluated at (x0 , y0 ) = (1, 1) for
∂x
∂y
p
f (x, y) = x x4 + y 2
Solution:
#
"
p
√
∂f
2x4
(1, 1) = p
+ x4 + y 2 (1, 1) = 2 2
∂x
x4 + y 2
"
#
∂f
xy
1
(1, 1) = p
(1, 1) = √
4
2
∂y
2
x +y
4. Find the directional derivative Du f (x0 , y0 ) for the function f (x, y) =
ex+y
, at the point (x0 , y0 ) =
+ y2
x2
(1, 2), where u = (u1 , u2 ) is any unit vector.
Solution: We have that
Du f (x0 , y0 ) = ∇f (x0 , y0 ) · u.
We compute
∇f (1, 2) =
(x2 + y 2 − 2x)ex+y x2 + y 2 − 2y)ex+y
,
(x2 + y 2 )2
(x2 + y 2 )2
|(x,y)=(1,2) = (
3e3 e3
, ).
25 25
Hence,
Du f (1, 2) =
e3
(3u1 + u2 ).
25
5. For f (x, y) = y 2 ln(x + e) and (x0 , y0 ) = (0, 1) find the equation of the tangent plane to the graph
z = f (x, y) at (x0 , y0 ).
Solution: The equation of the tangent plane for the function f at the point (x0 , y0 ) is given by
z = f (x0 , y0 ) +
∂f
∂f
(x0 , y0 )(x − x0 ) +
(x0 , y0 )(y − y0 ).
∂x
∂y
We compute
∂f
y2
1
(0, 1) =
|(0, 1) =
∂x
x+e
e
∂f
(0, 1) = 2y ln(x + e)|(0, 1) = 2
∂y
Since f (0, 1) = 1 we have that the equation of the tangent plane is given by:
z =1+
or
x
+ 2(y − 1)
e
x
+ 2y − z = 1.
e