Download 14 points - UCSD Math Department

NAME:
SID:
Midterm 2
Problem 1. (14 points)
i) (2 points) State the second fundamental theorem of calculus.
ii) (4 points) Using the second fundamental theorem of calculus compute
Z a(x)
d
f (t)dt.
dx b(x)
Prove your claim.
iii) (2 points) Write down the definition of pn (x), the Taylor polynomial
of f at x0 (assume f has enough derivatives).
iv) (2 points) Write down the main statement of the Lagrange remainder
theorem.
v) (2 points) Write down the formula for βk for β ∈ R and k ≥ 1. What
is the definition β0 ?
vi) (2 points) Write down Newton’s binomial expansion, that is the the
series expansion for (1 + x)β for β ∈ R and |x| < 1.
Rx
Solution. i) If f : [a, b] → R is continuous, then F (x) =
differentiable at all x ∈ (a, b). Moreover,
Z x
d
F 0 (x) =
f (t) dt = f (x).
dx a
a
f (t) dt is
ii)
d
dx
Z
a(x)
b(x)
d
f (t) dt =
dx
"Z
d
=
dx
"Z
a(x)
f (t) dt +
0
#
0
Z
f (t) dt
b(x)
a(x)
f (t) dt −
0
#
b(x)
Z
f (t) dt
0
= a0 (x)f (a(x)) − b0 (x)f (b(x)).
iii)
pn (x) =
n
X
f (k) (x0 )(x − x0 )k
k!
k=0
.
iv) Let f (x) and pn (x) be functions as in part iii). For each x, there exists
c (depending on x, x0 and n) between x and x0 such that
f (x) = pn (x) +
f (n+1) (c)(x − x0 )n+1
(n + 1)!
1
2
v)
β
(β)(β − 1) . . . (β − k + 1)
=
,
k
k!
β
=1
0
vi)
β
(1 + x) =
∞ X
β
k
k=0
xk
if x ∈ (−1, 1).
Problem 2. (12 points) Let f : R → R be a continuous function and
g : R → R be a differentiable function.
a) (4 points) Compute
Z x
d
g(x)f (t)dt.
dx 1
b) (8 points) Define
x
Z
(x − t)f (t)dt.
G(x) =
1
Prove that G00 (x) = f (x).
Solution. a) Note that
Z x
x
Z
g(x)f (t)dt = g(x)
f (t)dt
1
1
Using the product rule and the fundamental theorem of calculus
Z x
Z x
d
0
g(x)f (t)dt = g (x)
f (t)dt + g(x)f (x).
dx 1
1
b) We write
Z
x
Z
f (t)dt −
G(x) = x
1
x
tf (t)dt.
1
and use part a) to compute
Z x
Z
0
G =
f (t)dt + xf (x) − xf (x) =
1
x
f (t)dt.
1
Next, it follows G00 (x) = f (x).
Problem 3. (10 points) Use the Lagrange remainder theorem to prove
that:
1
x 2x2
x
1+ −
< (1 + x) 5 < 1 + , ∀x > 0
5
25
5
Solution. From the Lagrange remainder theorem, it follows
1
f (x) = (1 + x) 5 = 1 +
x f 00 (c) 2
+
x ,
5
2!
0 < c < x.
3
Since f 00 (c) =
1
5
9
9
4
· (− 45 )(1 + c)− 5 = − 25
(1 + c)− 5 and 0 < c < x we have
2x2
f 00 (c) 2
<
x < 0.
25
2!
from which the claimed inequality follows.
−
4
Problem 4. (14 points) Let f : R → R which has derivatives of all
orders. Assume that f 00 (x) = 2f 0 (x), ∀x ∈ R.
a) Find a recursive formula for the coefficients of the Taylor series of f at
x0 = 0.
b) Prove that f equals its Taylor series at x0 = 0 for all x ∈ R.
c) Consider now the Taylor series of f at x0 6= 0. Does f equal its Taylor
series for all x ∈ R?
Solution. a) From f 00 (x) = 2f 0 (x), ∀x ∈ R it is easy to see (by induction)
that, for n ≥ 2, f (n) (x) = 2f (n−1) (x), ∀x ∈ R and f (n) (x) = 2n−1 f 0 (x), ∀x ∈
R. We thus have, for n ≥ 2,
f (n) (0)
2f (n−1) (0)
2(n − 1)!an−1
2an−1
=
=
=
.
n!
n!
n!
n
b) Fix r ∈ R and denote I = [−r, r]. Since f (n) (x) = 2n−1 f 0 (x), ∀x ∈ R
we obtain
an =
|f (n) (x)| ≤ 2n−1 |f 0 (x)| ≤ 2n−1 sup |f 0 (x)| ≤ 2n−1 M
x∈I
|f 0 (x)|.
Using a theorem in the book we conclude that
where M = supx∈I
f equals its Taylor series on [−r, r] and since r is arbitrary, we obtain the
result on R.
c) The analysis above carries on for any other point x0 ∈ R, simply replace
I = [x0 − r, x0 + r].