Question

(1) Recall on February 6 in class we discussed e 0 + e 2πi/n + e 4πi/n + · · · + e 2(n−1)πi/n = 0 and in order to explain why it was true we needed to show that the sum of the real parts equals 0 and the sum of the imaginary parts is equal to 0.

(a) In class I showed the following identity for n even using the fact that sin(2π − x) = − sin(x): sin(0) + sin(2π/n) + sin(4π/n) + · · · + sin(2(n − 1)π/n) = 0 Do the same thing for n odd (make sure it is clear, at least to yourself, why the argument is slightly different for n even and n odd).

(b) Using the identity cos(x) = − cos(x + π), show that cos(0) + cos(2π/n) + cos(4π/n) + · · · + cos(2(n − 1)π/n) = 0 for n even.

(c) Why does the same proof not work for n odd ? Show and explain what goes wrong for the example of n = 3.

Answer #1

1) In the interval [0,2π) find all the solutions possible (in
radians ) :
a) sin(x)= √3/2
b) √3 cot(x)= -1
c) cos ^2 (x) =-cos(x)
2)The following exercises show a method of solving an equation
of the form: sin( AxB C + ) = , for 0 ≤ x < 2π . Find ALL
solutions .
d) sin(3x) = - 1/2
e) sin(x + π/4) = - √2 /2
f) sin(x/2 - π/3) = 1/2

If m>1 then x^m sin(1/x^n) will be differentiable at 0
However, why?
Q1. if we take limit x to 0 sin(1/x^n), we get sin(1/0) it
doesn't make sense even if m>1
Q2. why when M=1, it will not differentiable at 0?
Q3 Please tell me that limit x to infinity sin(1/x) DNE becasue
limit will be -1 and 1???? is that the reason?

1. Find the Fourier cosine series for f(x) = x on the interval 0
≤ x ≤ π in terms of cos(kx). Hint: Use the even extension.
2. Find the Fourier sine series for f(x) = x on the interval 0 ≤
x ≤ 1 in terms of sin(kπx). Hint: Use the odd extension.

14.) In Cantor's diagonalization, we construct a number x
between 0 and 1 that's not on the supposed list of real numbers
between 0 and 1. Recall, to construct x we make x's ith digit
(after the decimal point) equal to 1 if the corresponding digit of
the ith number on the list is even and we make x's ith digit 0
otherwise.
a) Suppose the list happens to start with the numbers 1/2, 1/3,
1/4, 1/5, 1/6, 1/7, 20.2/3,...

Recall from class that we defined the set of integers by
defining the equivalence relation ∼ on N × N by (a, b) ∼ (c, d) =⇒
a + d = c + b, and then took the integers to be equivalence classes
for this relation, i.e. Z = [(a, b)]∼ | (a, b) ∈ N × N . We then
proceeded to define 0Z = [(0, 0)]∼, 1Z = [(1, 0)]∼, − [(a, b)]∼ =
[(b, a)]∼, [(a, b)]∼...

1. a True or False? If ∫ [ f ( x ) ⋅ g ( x ) ] d x = [ ∫ f ( x )
d x ] ⋅ [ ∫ g ( x ) d x ]. Justify your answer.
B. Find ∫ 0 π 4 sec 2 θ tan 2 θ + 1 d θ
C. Show that ∫ 0 π 2 sin 2 x d x = ∫ 0 π 2 cos...

1. [10 marks] We begin with some mathematics regarding
uncountability. Let N = {0, 1, 2, 3, . . .} denote the set of
natural numbers.
(a) [5 marks] Prove that the set of binary numbers has the same
size as N by giving a bijection between the binary numbers and
N.
(b) [5 marks] Let B denote the set of all infinite sequences
over the English alphabet. Show that B is uncountable using a proof
by diagonalization.

1. [10] Let ~x ∈ R n with ~x 6= ~0. For each ~y ∈ R n , recall
that perp~x(~y) = ~y − proj~x(~y).
(a) Show that perp~x(~y + ~z) = perp~x(~y) + perp~x(~z) for all
~y, ~z ∈ R n .
(b) Show that perp~x(t~y) = tperp~x(~y) for all ~y ∈ R n and t ∈
R.
(c) Show that perp~x(perp~x(~y)) = perp~x(~y) for all ~y ∈ R
n

In class we proved that if (x, y, z) is a primitive Pythagorean
triple, then (switching x and y if necessary) it must be that (x,
y, z) = (m2 − n 2 , 2mn, m2 + n 2 ) for some positive integers m
and n satisfying m > n, gcd(m, n) = 1, and either m or n is
even. In this question you will prove that the converse is true: if
m and n are integers satisfying...

Let X1, . . . , Xn be i.i.d from pmf f(x|λ) where f(x) =
(e^(−λ)*(λ^x))/x!, λ > 0, x = 0, 1, 2
a) Find MoM (Method of Moments) estimator for λ
b) Show that MoM estimator you found in (a) is minimal
sufficient for λ
c) Now we split the sample into two parts, X1, . . . , Xm and
Xm+1, . . . , Xn. Show that ( Sum of Xi from 1 to m, Sum...

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