Monthly Archives: May 2011

Geometric Series Formulas

Let us consider the geometric series \( \sum_{k=t}^m ar^{k} \). The first \( n \) partial sums are

\(S_t = a r^t\)
\(S_{ t + 1 } = a r^t + a r^{ t + 1 }\)
\(S_{ t + 2 } = a r^t + a r^{ t + 1 } + a r^{ t + 2 }\)
\( \vdots \)
\(S_n = a r^t + a r^{ t + 1 } + a r^{ t + 2 } + \cdots + a r^{ t + j }\)

A formula for \( S_n\) by using the following cleverness; Factor \( r^t \) from the nth partial sum yielding

$$S_n = r^t \left( a + a r^t + a r^{ t + 1 } + \cdots + a r^{ j } \right)$$

which is equal to

$$S_n = r^t \left( \sum_{k=0}^{ j } ar^{k} \right) = \frac{ r^t \left( a – a r^{ j + 1} \right)}{ 1 – r }$$

Now consider the geometric series \( \sum_{k=t}^m ar^{k – 1} \). The first \( n \) partial sums are

\(S_t = a r^{ t – 1 }\)
\(S_{ t + 1 } = a r^{ t – 1 } + a r^t \)
\(S_{ t + 2 } = a r^{ t – 1 } + a r^t + a r^{ t + 1 }\)
\( \vdots \)
\(S_n = a r^{ t – 1 } + a r^t + a r^{ t + 1 } + \cdots + a r^{ t + j – 1 }\)

A formula for \( S_n\) by using the following cleverness; Factor \( r^t \) from the \( n \)th partial sum yielding

$$S_n = r^t \left( a + a r^t + a r^{ t + 1 } + \cdots + a r^{ j – 1 } \right)$$

which is equal to

$$ S_n = r^t \left( \sum_{k=1}^{ j } ar^{k – 1} \right) = \frac{ r^t \left( a – a r^{ j } \right)}{ 1 – r } $$

Integration by Partial Fractions Problems

This page contains solutions to the following problems

$$\int \! \frac{1}{x \left( M – x \right) } \, dx$$
$$\int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx$$
$$\int \! \frac{ \left( kx + l \right) }{ \left( ax + b \right) \left( cx + d \right) } \ dx$$

\( \displaystyle \int \! \frac{1}{x \left( M – x \right) } \ dx \)

Begin by decomposing the integrand.

$$ \frac{1}{x \left( M – x \right) } = \frac{A}{ x } + \frac{B}{ M – x } $$

Multiply through by \( x \left( M – x \right) \)

$$ 1 = A \left( M – x \right) + B x $$

If \( x = M \) then \( B = \frac{1}{ M } \)

If \( x = 0 \) then \( A = \frac{1}{ M } \)

therefore,

$$ \int \! \frac{1}{x \left( M – x \right) } \, dx = \frac{1}{M} \int \! \frac{1}{ x } \, dx + \frac{1}{M} \int \! \frac{1}{ M – x } \, dx $$
$$ \int \! \frac{1}{x \left( M – x \right) } = \frac{1}{M} \ln \left| x \right| – \frac{1}{M} \ln \left| M – x \right| + C = \frac{1}{M} \ln \left| \frac{x}{M – x} \right| + C$$

\( \displaystyle \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx \)

Decompose the integrand

$$ \frac{1}{ \left( ax + b \right) \left( cx + d \right) } = \frac{A}{ \left( ax + b \right)} \frac{B}{ \left( cx + d \right) }$$

Multiply through by \( \left( ax + b \right) \left( cx + d \right) \)

$$ 1 = A \left( cx + d \right) + B \left( ax + b \right) $$

If \( x = – \frac{d}{c} \), then \( 1 = B \left( a \left( -\frac{d}{c} \right) + b \right) \) yielding

$$ B = \frac{c}{ bc – ad } $$

Simimlarly,

If \( x = – \frac{b}{a} \), then \( 1 = A \left( c \left( -\frac{b}{a} \right) + d \right) \) and

$$ A = \frac{a}{ ad – bc } $$

substituting

$$ \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx = \frac{a}{ ad – bc } \int \! \frac{1}{ \left( ax + b \right) } \, dx + \frac{c}{ bc – ad } \int \! \frac{1}{ \left( cx + d \right) } \, dx $$

\( p = a x + b \ \ \ \ \ \ dp = a \ dx \)

\( q = c x + d \ \ \ \ \ \ dq = c \ dx \)

$$ \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \, dx = \frac{a}{ ad – bc } \frac{1}{a} \int \! \frac{1}{ p } \, dx + \frac{c}{ bc – ad } \frac{1}{c} \int \! \frac{1}{ q } \, dx $$
$$ \int \! \frac{1}{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{1}{ ad – bc } \ln \left( ax + b \right) + \frac{1}{bc – ad} \ln \left( c x + d \right) + C = \frac{1}{ad-bc} \ln \left( \frac{ a x + b }{ c x + d } \right) + C $$

\( \displaystyle \int \! \frac{ \left( kx + l \right) }{ \left( ax + b \right) \left( cx + d \right) } \ dx \)

\( \frac{ k x + l }{ \left( ax + b \right) \left( cx + d \right) } = \frac{A}{ \left( ax + b \right)} \frac{B}{ \left( cx + d \right) }\)

\( k x + l = A \left( cx + d \right) + B \left( ax + b \right) \)

\( x = – \frac{d}{c} \)

\( k \left( – \frac{d}{c} \right) + l = B \left( a \left( -\frac{d}{c} \right) + b \right) \)

\( B = \frac{ c l – d k }{ bc – ad } \)

\( x = – \frac{b}{a} \)

\( k \left( – \frac{b}{a} \right) + l = A \left( c \left( -\frac{b}{a} \right) + d \right) \)

\( A = \frac{ a l – k b }{ ad – bc } \)

\( \int \! \frac{ \left( kx + l \right) }{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{ a l – k b }{ ad – bc } \int \! \frac{1}{ a x + b } \ dx + \frac{ c l – d k }{ bc – ad } \int \! \frac{1}{ c x + d } \ dx \)

\( p = a x + b \ \ \ \ \ \ dp = a \ dx \)

\( q = c x + d \ \ \ \ \ \ dq = c \ dx \)

\( \int \! \frac{ k x + l }{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{a l – k b}{ ad – bc } \frac{1}{a} \int \! \frac{1}{ p } \ dx + \frac{ c l – d k }{ bc – ad } \frac{1}{c} \int \! \frac{1}{ q } \ dx \)

\( \int \! \frac{ k x + l}{ \left( ax + b \right) \left( cx + d \right) } \ dx = \frac{a l – k b}{ a^2 d – abc } \ln \! \left( ax + b \right) + \frac{ c l – d k}{bc^2 – acd} \ln \! \left( c x + d \right) + C = \frac{ a l – k b}{ a^2 d – abc } \ln \! \left( ax + b \right) + \frac{ d k – c l }{acd – bc^2} \ln \! \left( c x + d \right) + C \)

Integration by Parts Problems

$$ \int \! u \ dv = uv \ – \int \! v \ du$$

\( \int \! \ln x \ dx \)

\( u = \ln x \ \ \ \ \ \ du = \frac{1}{x} \ dx \)

\( dv = dx \ \ \ \ \ \ v = x \)

\( \int \! \ln x \ dx = x \ln x – \int \! x \frac{1}{x} \ dx \)

\( \int \! \ln x \ dx = x \ln x – x + C \)

 

\( \int \! x \ln x \ dx \)

\( u = \ln x \ \ \ \ \ \ du = \frac{1}{x} \ dx \)

\( dv = x \ dx \ \ \ \ \ \ v = \frac{x^2}{2}\)

\( \int \! x \ln x \ dx = \frac{x^2}{2} \ln x – \int \! \frac{x^2}{2} \frac{1}{x} \ dx \)

\( \int \! x \ln x \ dx = \frac{x^2}{2} \ln x – \frac{x^2}{4} + C \)

 

\( \int \! \sin^{-1} x \ dx \)

\( u = \sin^{-1} x \ \ \ \ \ \ du = \frac{1}{\sqrt{1 – x^2}} \ dx\)

\( dv = dx \ \ \ \ \ \ v = x \)

\( \int \! \sin^{-1} x \ dx = x \sin^{-1} x – \int \! x \frac{1}{ \sqrt{1 – x^2 }} \ dx\)

\( \int \! \frac{x}{ \sqrt{1 – x^2 }} \ dx\)

\( t = 1 – x^2 \ \ \ \ \ \ – \frac{1}{2} \ dt = x \ dx\)

\( – \int \! \frac{1}{ 2 \sqrt{ t }} \ dt = – \sqrt{ t } + C\)

\( \int \! \sin^{-1} x \ dx = x \sin^{-1} x – \int \! x \frac{1}{ \sqrt{1 – x^2 }} \ dx = x \sin^{-1} x + \sqrt{ 1 – x^2 } + C \)

 

\( \int \! x e^{ x } \ dx \)

\( u = x \ \ \ \ \ \ du = dx \)

\( dv = e^{x} \ dx \ \ \ \ \ \ v = e^{x}\)

\( \int \! x e^{ x } \ dx = x e^{x} – \int \! e^{x} \ dx = x e^{x} – e^{x} + C\)

 

\( \int \! e^{ \sqrt{x} } \ dx \)

\( t = \sqrt{x} \ \ \ \ \ \ dt = \frac{1}{2 \sqrt{ x }} \ dx\)

\( 2 t \ dt = dx \)

\( \int \! 2 t e^{ t } \ dt \)

\( u = 2 t \ \ \ \ \ \ du = 2 \ dt \)

\( dv = e^{t} \ dt \ \ \ \ \ \ v = e^{t}\)

\( \int \! 2 t e^{ t } \ dt = 2 t e^{t} – \int \! 2 e^{t} \ dt = 2 t e^{t} – 2 e^{t} + K\)

\( \int \! e^{ \sqrt{x} } \ dx = 2 \sqrt{ x } e^{ \sqrt{ x } } -2 e^{ \sqrt{ x } } + C\)

 

\( \int \! e^{ x } \sin x \ dx \)

\( u = \sin x \ \ \ \ \ \ dv = e^{ x } \ dx\)

\( du = \cos x \ dx \ \ \ \ \ \ v = e^{ x } \)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – \int \! e^{ x } \cos x \ dx \)

\( s = \cos x \ \ \ \ \ \ ds = – \sin x \ dx \)

\( dt = e^{x} \ dx \ \ \ \ \ \ t = e^{x}\)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – \left( e^{ x } \cos x – \int \! – e^{ x } \sin x \ dx \right)\)

\( \int \! e^{ x } \sin x \ dx = e^{ x } \sin x – e^{ x } \cos x – \int \! e^{ x } \sin x \ dx \)

\( \int \! e^{ x } \sin x \ dx = \frac{1}{2} \left( e^{ x } \sin x – e^{ x } \cos x \right) + C \)

Set Theory Problems

Let (a, b, c, d in mathbb{R} ) be objects not necessarily distinct from one another; moreover, let ( A = { {a } , { a, b } } ) and let ( B = { {c } , {c , d } } ). Prove that ( A = B ) if and only if ( a = c ) and ( b = d ).

To prove the above theorem requires proving the following two statements:
(i) If ( A = B ) then ( a = c ) and ( b = d ).
(ii) If ( a = c ) and ( b = d ) then ( A = B ).

By definition the two sets ( A ) and ( B ) are equal if every element in ( A ) is also in ( B ) and if every element in ( B ) is also in ( A ). Let ( P = { a } ), ( Q = {a , b } ), ( R = { c } ), and ( R = {c , d } ). Now consider statement (i). If ( A = B ) then ( P,Q in B ) and ( R,S in A ) which implies ( P = R ) and ( Q = S ) which by definition implies ( a = c ) and ( b = d ). Now consider (ii). If ( a = c ) and ( b = d ) then by definition ( P = R ) and ( Q = S ). Since ( P = R ) and ( Q = S ) it is the case that ( A = B ) since every element in ( A ) is also in ( B ) and every element in ( B ) is also in ( A ).