Keshav Srinivasan
Spruck
Math 605 Homework 4
1. A. Suppose that f is measurable. Then
f k f is measurable.
k
k, so
f
k is measurable for each k, so
By theorem 4.1,
f k f is measurable.
Since
f
f
k
Lab J: Synthesis of Triphenylmethanol from a Grignard
Reaction
In designing new molecules, it is usually necessary to form new carboncarbon bonds. The Grignard reaction represents one such carbon-carb
Lab D: The Synthesis of Isopentyl Acetate (Banana
Oil)
Laboratory Techniques:
Refluxing a reaction
Organic Chemistry Concepts:
Esterification of carboxylic acids
In this lab, you will use a Fischer es
1
2
Lab F: Synthesis of Dimethyl 3,4,5,6Tetraphenylphthalate from a Diels Alder Reaction
The ability to form carbon-carbon bonds is an important goal in organic
chemistry. We will employ a very famous
Lab E: Aldol Condensation: Synthesis of 2,3,4,5Tetraphenylcyclopentadienone
The ability to form carbon-carbon bonds is an important technique in
organic chemistry. In the Grignard Lab, we formed C-C b
Lab G: Aerobic Oxidation of p-Bromobenzyl Alcohol using a
Cu(I)/TEMPO Catalyst
Courtesy of:
Hill, N.J., Hoover, J.M., Stahl, S.S. J. Chem. Educ. 2013, 90, 102-105.
Please see your article packet for b
Final exam solutions
1. Step 1. Assume rst |E| < . Then := sup |EI| < , where the sup is
|I|
taken over countable coverings by nonempty open intervals. It suces to show = 1.
Clearly 1 so we need to sh
REAL VARIABLES: PROBLEM SET 1
BEN ELDER
1. Problem 1.1a
First lets prove that limsup Ek consists of those points which belong to innitely many
Ek . From equation 1.1:
j=1
k=j
limsup Ek =
Ek
For limsup
Real Variables: Solutions to Homework 3
September 23, 2011
Exercise 0.1. Chapter 3, # 2: Show that the cantor set C consists of all x such that x has
some triadic expansion for which every ck is eithe
Real Variables: Solutions to Homework 2
September 18, 2011
Exercise 0.1. Chapter 2, # 1: Let f (x) = x sin(1/x) for x (0, 1] and f (0) = 0. Show
that f is bounded and continuous on [0, 1] but V [f ; 0
REAL VARIABLES: PSET 7
1. Problem 6.1
a) Consider E . We know that Ex = cfw_y : (x, y) E has measure 0 in R1 . Theorem 5.11
says that the integral of a non-negative function is zero only if the functi
Keshav Srinivasan
Spruck
Math 605 Homework 5
1. Since f is measurable,
f
exists. Since all the Ek are disjoint, by theorem 5.24 we have
E
f =
k
E
f = f
k
Ek
Ek
= ak = ak Ek .
k
Ek
k
Ek
2. If f k = (
REAL VARIABLES: PSET 10
1. Problem 8.10
Using Lemma 3.22, we can approximate E from within by closed sets. Choose a sequence
1
k = k . Then for every k , there is a set Fk such that k+1 < |E Fk | < k
Real Variables: Solutions to Homework 9
Theodore D. Drivas
November 11, 2011
Exercise 0.1. Chapter 8, # 1: For complex-valued, measurable f , f = f1 + if2 with fi
real-valued and measurable, we have E
REAL VARIABLES PSET 6
1. Problem 5.10
f is in Lp (E) because |f |p 2p (|f fk |p + |fk |p ), and the right side of this equality is
bounded.
From Theorem 5.51, we know that:
|f |p = p
0
E
p
|f fk | +
E
Labs H and I: Synthesis of Acetyl Ferrocene and
Purification Using Column Chromatography
Cyclopentadiene is a somewhat unusual hydrocarbon molecule in that its
C-H bond has a pKa of ~15. This is due t