TuesEx1.) det[ 3x3 matrix with elements M[ij] ] write or check that
det(M) = det(M)
so I guess using a 3x3 matrix show that the determinant of the
transposed matrix and the determinant of the original matrix are equal
TuesEx2.) Show that: (ab) a = 0 = (ab
The Bolzano-Weierstrass Property and Compactness
Denitions: Let S be a subset of R. An upper bound of S is a number b such that x
for every x S . A least upper bound of S is a b such that b
b
b for every b thats an
upper bound of S .
Remark: In R, a set c
Math 347
Epsilonics IV: Sequences and Limits, II Solutions
A.J. Hildebrand
Practice Problems: Four Classic Proofs Solutions
1. Proof of the Monontone Convergence Theorem from the Completeness Axiom: Using the denition of a sup and the Completeness Axiom,
Definitions: Convergent and divergent series
1. A series is an infinite sum, written
ak . It doesnt have to start at k = 0.
k =0
2. The nth partial sum of the series is Sn = a0 + a1 + a2 + . + an
3. The series
ak
converges to a number L means: lim Sn =
Nested Interval Theorem
Let cfw_[an , bn ] be a sequence of closed intervals such that n R, [an+1 , bn+1 ][an , bn ]. and
that limn bn an 0. Then by the Nested Interval Theorem,
In =
(1)
n=1
and is consisted of a single point.
Monotone Convergence Theore
Nested Interval Theorem
Let cfw_[an , bn ] be a sequence of closed intervals such that n R, [an+1 , bn+1 ][an , bn ]. and
that limn bn an 0. Then by the Nested Interval Theorem,
In =
(1)
n=1
and is consisted of a single point.
Monotone Convergence Theore
Proof of the MCT using the NIT
Nested Interval Theorem
Let cfw_[an , bn ] be a sequence of closed intervals such that n R, [an+1 , bn+1 ][an , bn ]. and
that limn bn an 0. Then by the Nested Interval Theorem,
In =
(1)
n=1
and is consisted of a single poi
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TRIGONOMETRIC IDENTITIES
Reciprocal identities
1
1
sin u =
cos u =
csc u
sec u
1
1
tan u =
cot u =
cot u
tan u
1
1
csc u =
sec u =
sin u
cos u
Pythagorean Identities
Power-Reducing/Half Angle Formulas
1 cos(
Common Derivatives and Integrals
Derivatives
Basic Properties/Formulas/Rules d ( cf ( x ) ) = cf ( x ) , c is any constant. ( f ( x ) g ( x ) ) = f ( x ) g ( x ) dx dn d ( c ) = 0 , c is any constant. ( x ) = nxn-1 , n is any number. dx dx f f g - f g (Qu
Math 347
Epsilonics IV: Sequences and Limits, II Solutions
A.J. Hildebrand
Practice Problems: Four Classic Proofs Solutions
1. Proof of the Monontone Convergence Theorem from the Completeness Axiom: Using the denition of a sup and the Completeness Axiom,
The Bolzano-Weierstrass Property and Compactness
Denitions: Let S be a subset of R. An upper bound of S is a number b such that x
for every x S . A least upper bound of S is a b such that b
b
b for every b thats an
upper bound of S .
Remark: In R, a set c