SUGGESTIONS FOR ANALYSIS II - SERIES 16
Exercise 61
Let us consider a differential equation of the form
0
y = f (x)g(y)
y(x0 ) = y0
To find a solution we use the general formula
Z x
Z y
du
f (s) d s
=
x0
y0 g(u)
and we try to explicit y as function of x. Pay attention to the correct domain of
definition of f and g.
Exercise 62
• To draw a graphic of φ we just study the given function. To prove that
φ is a C ∞ function, we notice that φ is of class C ∞ in the open intervals
(−∞, 0) and (0, +∞). To check that φ is of class C ∞ also at t = 0 we just
compute the derivatives of φ on the open intervals (−∞, 0) and (0, +∞)
and then we take te limit in both sides as t → 0. To be more precise, for
each k ≥ 0 we have
φ(k) (t) = 0
(t) − 1
φ(k) (t) = ptk2k
e t
if t ∈ (−∞, 0)
if t ∈ (0, +∞)
where pk (t) is a polynomial in t with real coefficients. This implies that
φ(k) (t) goes to 0 as t → 0+ . On the other hand, φ(k) (t) is constantly 0 for
t < 0. So that,
lim φ(k) (t) = 0
t→0
and this proves that φ is of class C k for each k ≥ 0. Thus, φ is of class C ∞ .
• It is enough to set
1
ak = R +∞
φ(1 − k 2 x2 ) d x
−∞
and noticing that
supp(φk ) = {x ∈ R | φk (x) 6= 0} ⊆ [−k.k]
Exercise 63
First of all we recall the following formula
Z π
1
1
i(d−k)x
(1)
e
dx =
0
2π −π
if d = k
if d =
6 k
Thanks to this, for |k| ≤ n we find
!
Z π
Z π
n
X
1
1
−ikx
idx
fˆ(k) =
f (x)e
dx =
cd e
e−ikx d x =
2π −π
2π −π
d=−n
Z π
n
X
1
=
cd
ei(d−k)x d x = ck
2π −π
d=−n
and similarly we deduce that fˆ(k) = 0 for |k| > 0.
1
2
SUGGESTIONS FOR ANALYSIS II - SERIES 16
To prove the Parseval identity we recall (1) and the fact that
ck if |k| ≤ n
ˆ
f (k) =
0 if |k| > n
in order to write

!
!
Z π
Z π
X
X
1
1

|f (x)|2 d x =
cd eidx
ck eikx  d x =
2π −π
2π −π
d∈Z
k∈Z



Z π
X
X
1


cd eidx  
=
ck eikx  d x =

2π −π
−n≤d≤n
=
X
−n≤k≤n
cd ck fˆ(d)fˆ(k)
−n≤d,k≤n
=
X
−n≤k≤n
|ck |2 =
X
1
2π
Z
π
e
i(d−k)x
dx
=
−π
|fˆ(k)|2
k∈Z
Exercise 64
• From Newton binomial formula we now that
n X
n k n−k
a b
(a + b)n =
k
k=0
We use this hint and write 1 = x + (1 − x) to get the desired result.
• Taking partial derivatives with respect to a in the Newton binomial formula
we get
n
X
n(a + b)n−1 =
kak−1 bn−k
k=0
and then, if we multiplicate by a both sides in the previous equation we
find
n
X
n−1
na(a + b)
=
kak bn−k
k=0
So that, to prove the given statement it is enough to substitute a = x and
b = 1 − x. For the second formula we just need to take the second partial
derivative with respect to a in the Newton binomial formula and finally
substitute a = x and b = 1 − x.
• Look at Section 3 of the given side notes.