Partial Derivatives
Let f be a function of x and y. Suppose that f
is dened in a neighborhood of (x0 , y0 ). The partial
derivative of f with respect to x at (x0 , y0 ) is
f (x0 + h, y0 ) f (x0 , y0 )
f
(x0 , y0 ) = lim
h0
x
h
if the limit exists. Similar
Parametric Equations
A curve C in plane can be represented by parametric equations
x = f (t), y = g(t),
t I,
where f and g are functions on an interval I. Each
value of t determines a point (x, y) = (f (t), g(t) in
the plane. As t varies over I, the point
Polar Coordinates
We choose a point O in the plane called the
pole (or origin). Then we draw a ray starting at
O, called the polar axis. If P is any other point
in the plane, let r be the distance from O to P and
let be the angle between the polar axis an
Solution 2
Bahman Gharesifard1
Solution 1. The answer to each question is provide below.
(i) The function is clearly continuous. A simple inductive argument, similar to the one used in Homework 1, shows that for x > 0, the kth derivative of g is
g( k) ( x
Solution 1
Bahman Gharesifard1
Solution 1.
(i) [10] This is a direct calculation. We have
t
s
t
1
e RC e RC x(s a)ds
RC
t a
t
s+ a
1
s
e RC e RC x(s)d
=
RC
t a (t a)+s
1
s
e RC x(s)d
=
RC
= a ( Ax(t)
Aa x(t) =
(ii) [10] Similar to what we did in clas
Solution 3
Bahman Gharesifard1
Solution 1. We prove the claims:
(i) Since f , g S (R ), we have that f , g C (R ) and hence f g C (R ). For q Z 1 , we have
( f g) ( q ) =
q
k=0
q ( k) ( p k)
f g
.
k
Therefore, for all p, q Z 1 , we have
q
lim | x| p |( f
Solution 4
Bahman Gharesifard1
Solution 1. (Note: In the denition of Fourier transform that I used, I ignored the 2, which is ne.
So please adjust for that in the solution; your solutions will have this 2 carried forward everywhere.)
Now, it follows from
Solution 5
Bahman Gharesifard1
Solution 1. Since T S (R ) is a periodic distribution with period a > 0, in the sense of S (R ), we
have
+
t
n e2in a .
T=
n =
Hence, for all S (R ), we have that
+
T, =
t
n e2in a ,
n =
+
=
t
n e2in a ,
n =
+
=
t
n
n =