Tag Archives: series

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

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

Lower Darboux Sum

If \( f : \left[ a, \ b \right] \rightarrow \mathbb{R} \) is a bounded function then for an arbitrary partition of \(\left[ a, \ b \right]\) let

$$m = \inf_{\left[ a, \ b \right]} f \left( x \right)$$

and

$$m_i = \inf_{\left[ x_i, \ x_{i-1} \right]} f \left( x \right)$$

The Lower Darboux sum is defined as

$$L \left( f, \ P \right) = \sum_{i=1}^n m_i \left( x_i – x_{i-1} \right)$$

Note that \(f\) needs to be defined on \(\left[ a, \ b \right]\) but it does not have to be continuous. The following diagrams illustrate possible Lower Darboux sums. The diagrams should be understood as representative but not as definitive.

graphics coming soon

Notice that if \(P\) is any partition of \( \left[ a, \ b \right] \) then \( m \left( b-a \right) \le L \left( f, \ P \right) \). By definition \( m \le m_i \) therefore,

$$m \sum_{i=1}^n \left( x_i – x_{i-1} \right) = \sum_{i=1}^n m \left( x_i – x_{i-1} \right) \le \sum_{i=1}^n m_i \left( x_i – x_{i-1} \right)$$

it is the case that

$$\sum_{i=1}^n \left( x_i – x_{i-1} \right) = \left( x_1 – x_0 \right) + \left( x_2 – x_1 \right) + \left( x_3 – x_2 \right) + \dots + \left( x_{n-1} – x_{n-2} \right) + \left( x_n – x_{n-1} \right) = \left( x_n – x_0 \right) = \left( b – a \right)$$

therefore, \( m \left( b-a \right) \le L \left( f, \ P \right) \) as claimed.

If the partition \( P \) of the above graphics are refined to a partition \( R \) then

graphics coming soon

From the above diagrams it seems reasonable to suggest \( L \left( f, \ P \right) \le L \left( f, \ R \right) \).

Consider an arbitrary interval in \( P \), \( \left[ x_{i-1}, \ x_i \right] \) refined into \( \left[ x_{i-1}, \ t \right] \cup \left[ t, \ x_i \right] \). Let

$$m_{t^{-}} = \inf_{\left[ x_{i-1}, \ t \right]} f \left( x \right)$$
$$m_{t^{+}} = \inf_{\left[ t, \ x_i \right]} f \left( x \right)$$

By definition

$$m_i \le m_{t^{-}}$$
$$m_i \le m_{t^{+}}$$

The area of the refinement of is

$$m_i \left( x_i – x_{i-1} \right) = m_i \left( t – x_{i-1} \right) + m_i \left( x_i – t \right) \le m_{t^{-}} \left( t – x_{i-1} \right) + m_{t^{+}} \left( x_i – t \right)$$

Hence, reasoning inductively it is indeed true that

$$L \left( f, \ P \right) \le L \left( f, \ R \right)$$

Upper Darboux Sum

If \( f : \left[ a, \ b \right] \rightarrow \mathbb{R} \) is a bounded function then for an arbitrary partition of \(\left[ a, \ b \right]\) let

$$M = \sup_{\left[ a, \ b \right]} f \left( x \right)$$

and

$$M_i = \sup_{\left[ x_{i-1}, \ x_i \right]} f \left( x \right)$$

The Upper Darboux sum is defined as

$$U \left( f, \ P \right) = \sum_{i=1}^n M_i \left( x_i – x_{i-1} \right)$$

Note that \(f\) needs to be defined on \(\left[ a, \ b \right]\) but it does not have to be continuous. The following diagrams illustrate possible Upper Darboux sums. The diagrams should be understood as representative but not as definitive.

graphics coming soon

Notice that if \(P\) is any partition of \( \left[ a, \ b \right] \) then \( U \left( f, \ P \right) \le M \left( b-a \right) \). By definition \( M_i \le M \) therefore,

$$\sum_{i=1}^n M_i \left( x_i – x_{i-1} \right) \le \sum_{i=1}^n M \left( x_i – x_{i-1} \right) = M \sum_{i=1}^n \left( x_i – x_{i-1} \right)$$

it is the case that

$$\sum_{i=1}^n \left( x_i – x_{i-1} \right) = \left( x_1 – x_0 \right) + \left( x_2 – x_1 \right) + \left( x_3 – x_2 \right) + \dots + \left( x_{n-1} – x_{n-2} \right) + \left( x_n – x_{n-1} \right) = \left( x_n – x_0 \right) = \left( b – a \right)$$

therefore, \( U \left( f, \ P \right) \le M \left( b-a \right) \) as claimed.

If the partition \( P \) of the above graphics are refined to a partition \( R \) then

graphics coming soon

From the above diagrams it seems reasonable to suggest \( U \left( f, \ R \right) \le U \left( f, \ P \right) \).

Consider an arbitrary interval in \( P \), \( \left[ x_{i-1}, \ x_i \right] \) refined into \( \left[ x_{i-1}, \ t \right] \cup \left[ t, \ x_i \right] \). Let

$$M_{t^{-}} = \sup_{\left[ x_{i-1}, \ t \right]} f \left( x \right)$$
$$M_{t^{+}} = \sup_{\left[ t, \ x_i \right]} f \left( x \right)$$

By definition

$$M_{t^{-}} \le M_i$$
$$M_{t^{+}} \le M_i$$

The area of the refinement is

$$M_{t^{-}} \left( t – x_{i-1} \right) + M_{t^{+}} \left( x_i – t \right) \le M_i \left( t – x_{i-1} \right) + M_i \left( x_i – t \right) = M_i \left( x_i – x_{i-1} \right)$$

Hence, reasoning inductively it is indeed true that

$$U \left( f, \ R \right) \le U \left( f, \ P \right)$$

\( \sin x \) centered about \( \frac{ \pi }3 \)

The Taylor Series for \( \sin x \) centered at \( x=\frac{\pi}3 \)

\( g(x)=\sin x \) \( g^{\prime}\left(\frac{\pi}3 \right)=\frac{1}2 \)
\( g^{\prime}(x)=\cos x \) \( g^{\prime}\left(\frac{\pi}3 \right)=\frac{1}2 \)
\( g^{\prime\prime}(x)=-\sin x \) \( g^{\prime\prime}\left(\frac{\pi}3 \right)=\frac{-\sqrt 3}2 \)
\( g^{\prime\prime\prime}(x)=-\cos x \) \( g^{\prime\prime\prime}\left(\frac{\pi}3 \right)=\frac{-1}2 \)
\( g^{(4)} (x)=\sin x \) \( g^{(4)}\left(\frac{\pi}3 \right)=\frac{\sqrt 3}2 \)
\(\sin x =\frac {\sqrt 3}{2 \cdot 0!}\left(x-\frac{\pi}3 \right)^{0}+\frac {1}{2\cdot 1!}\left(x-\frac {\pi}3 \right)^{1}-\frac {\sqrt 3}{2 \cdot 2!}\left(x-\frac {\pi}3 \right)^{2}-\frac {1}{2 \cdot 3!} \left(x- \frac{\pi}3 \right)^{3}+\frac {\sqrt 3}{2 \cdot 4!} \left(x- \frac{\pi}3 \right)^{4}+ \cdots\)
\(\sin x =\frac{\sqrt 3}{2} \left(1-\frac{1}{2!} \left(x-\frac{\pi}3 \right)^{2}+\frac{1}{4!} \left(x-\frac{\pi}3 \right)^{4}- \cdots \right)+ \frac{1}2 \left( \left( x – \frac{\pi}3 \right) – \frac{1}{3!} \left( x – \frac{\pi}3 \right)^{3} + \frac{1}{5!} \left( x – \frac{\pi}3 \right)^{5} + \cdots \right)\)
\(\sin x=\frac {\sqrt 3}2 \sum_{n=0}^{\infty} \frac{ \left(-1 \right)^n}{ \left(2n \right)!} \left( x – \frac {\pi}3 \right)^{2n}+ \frac {1}2 \sum_{n=0}^{\infty} \frac { \left(-1 \right)^{n}}{ \left(2n+1 \right)!} \left( x – \frac{\pi}3 \right)^{2n+1}\)