Exponent shift method for power series summation - Revision history

Vipul: /* Examples */

2012-07-06T22:06:40Z

Examples

← Older revision		Revision as of 22:06, 6 July 2012
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	\| <math>\sum_{k=0}^\infty \frac{x^k}{(k + 2)!}</math> \|\| Shifted index of summation + missing terms \|\| <math>\frac{1}{x^2}\sum_{l=2}^\infty \frac{x^l}{l!} = \frac{1}{x^2} \left[\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right) - \left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)\right]</math> \|\| <math>\frac{e^x - 1 - x}{x^2}</math>, expression valid for <math>x \ne 0</math>. At <math>x = 0</math>, we get a value of <math>1/2</math>.		\| <math>\sum_{k=0}^\infty \frac{x^k}{(k + 2)!}</math> \|\| Shifted index of summation + missing terms \|\| <math>\frac{1}{x^2}\sum_{l=2}^\infty \frac{x^l}{l!} = \frac{1}{x^2} \left[\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right) - \left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)\right]</math> \|\| <math>\frac{e^x - 1 - x}{x^2}</math>, expression valid for <math>x \ne 0</math>. At <math>x = 0</math>, we get a value of <math>1/2</math>.
	\|-		\|-
	\| <math>\sum_{k=0}^\infty \frac{x^k}{k + 1}</math> \|\| Shifted index of summation \|\| <math>\frac{1}{x} \sum_{l=1}^\infty \frac{x^l}{l} \|\| <math>\frac{-\ln(1 - x)}{x}</math> for <math>x \in [-1,0) \cup (0,1)</math>, 1 for <math>x = 0</math>.		\| <math>\sum_{k=0}^\infty \frac{x^k}{k + 1}</math> \|\| Shifted index of summation \|\| <math>\frac{1}{x} \sum_{l=1}^\infty \frac{x^l}{l}</math> \|\| <math>\frac{-\ln(1 - x)}{x}</math> for <math>x \in [-1,0) \cup (0,1)</math>, 1 for <math>x = 0</math>.
	\|}		\|}

Vipul: /* Examples */

2012-07-06T22:04:54Z

Examples

← Older revision		Revision as of 22:04, 6 July 2012
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	\| <math>\sum_{k=0}^\infty \frac{x^k}{(k + 2)!}</math> \|\| Shifted index of summation + missing terms \|\| <math>\frac{1}{x^2}\sum_{l=2}^\infty \frac{x^l}{l!} = \frac{1}{x^2} \left[\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right) - \left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)\right]</math> \|\| <math>\frac{e^x - 1 - x}{x^2}</math>, expression valid for <math>x \ne 0</math>. At <math>x = 0</math>, we get a value of <math>1/2</math>.		\| <math>\sum_{k=0}^\infty \frac{x^k}{(k + 2)!}</math> \|\| Shifted index of summation + missing terms \|\| <math>\frac{1}{x^2}\sum_{l=2}^\infty \frac{x^l}{l!} = \frac{1}{x^2} \left[\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right) - \left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)\right]</math> \|\| <math>\frac{e^x - 1 - x}{x^2}</math>, expression valid for <math>x \ne 0</math>. At <math>x = 0</math>, we get a value of <math>1/2</math>.
			\|-
			\| <math>\sum_{k=0}^\infty \frac{x^k}{k + 1}</math> \|\| Shifted index of summation \|\| <math>\frac{1}{x} \sum_{l=1}^\infty \frac{x^l}{l} \|\| <math>\frac{-\ln(1 - x)}{x}</math> for <math>x \in [-1,0) \cup (0,1)</math>, 1 for <math>x = 0</math>.
	\|}		\|}

Vipul: /* Examples */

2012-07-06T21:48:58Z

Examples

← Older revision		Revision as of 21:48, 6 July 2012
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	\| <math>\sum_{k=1}^\infty \frac{x^k}{k!}</math> \|\| Missing terms only \|\| <math>\left(\sum_{k=0}^\infty \frac{x^k}{k!} \right) - \frac{x^0}{0!}</math> \|\| <math>e^x - 1</math>		\| <math>\sum_{k=1}^\infty \frac{x^k}{k!}</math> \|\| Missing terms only \|\| <math>\left(\sum_{k=0}^\infty \frac{x^k}{k!} \right) - \frac{x^0}{0!}</math> \|\| <math>e^x - 1</math>
	\|0
	\|-		\|-
	\| <math>\sum_{k=0}^\infty \frac{x^k}{(k + 2)!}</math> \|\| Shifted index of summation + missing terms \|\| <math>\frac{1}{x^2}\sum_{l=2}^\infty \frac{x^l}{l!} = \frac{1}{x^2} \left[\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right) - \left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)\right]</math> \|\| <math>\frac{e^x - 1 - x}{x^2}</math>, expression valid for <math>x \ne 0</math>. At <math>x = 0</math>, we get a value of <math>1/2</math>.		\| <math>\sum_{k=0}^\infty \frac{x^k}{(k + 2)!}</math> \|\| Shifted index of summation + missing terms \|\| <math>\frac{1}{x^2}\sum_{l=2}^\infty \frac{x^l}{l!} = \frac{1}{x^2} \left[\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right) - \left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)\right]</math> \|\| <math>\frac{e^x - 1 - x}{x^2}</math>, expression valid for <math>x \ne 0</math>. At <math>x = 0</math>, we get a value of <math>1/2</math>.
	\|}		\|}

Vipul: /* Shifted index of summation */

2012-07-06T21:47:42Z

Shifted index of summation

← Older revision		Revision as of 21:47, 6 July 2012
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	Let's now consider another summation, with <math>k_1, r</math> integers such that <math>k_1 + r \ge k_0</math>:		Let's now consider another summation, with <math>k_1, r</math> integers such that <math>k_1 + r \ge k_0</math>:

	<math>\sum_{k=k_1}^\infty a_{k+r}x^k/math>		<math>\sum_{k=k_1}^\infty a_{k+r}x^k</math>

	We simplify this as follows		We simplify this as follows

Vipul: Created page with "==Description of the method== This method is used for summing up various power series that look similar to existing power series but have some terms missing, extra, or shifte..."

2012-07-06T21:46:33Z

Created page with "==Description of the method== This method is used for summing up various power series that look similar to existing power series but have some terms missing, extra, or shifte..."

New page

==Description of the method==

This method is used for summing up various power series that look similar to existing power series but have some terms missing, extra, or shifted. It is closely related to another method, the [[substitution method for power series summation]].

===Missing terms in summation===

Suppose we know how to do a certain summation <math>\sum_{k=k_0}^\infty a_kx^k</math>, and we are asked to do a summation <math>\sum_{k=k_1}^\infty a_kx^k</math>, where <math>k_1 \ge k_0</math> is a positive integer. This is given by:

<math>\sum_{k = k_1}^\infty a_kx^k = \sum_{k=k_0}^\infty a_kx^k - \sum_{k=k_0}^{k_1 - 1} a_kx^k</math>

===Shifted index of summation===

Suppose we know how to do a summation of the form:

<math>\sum_{k=k_0}^\infty a_kx^k</math>

Let's now consider another summation, with <math>k_1, r</math> integers such that <math>k_1 + r \ge k_0</math>:

<math>\sum_{k=k_1}^\infty a_{k+r}x^k/math>

We simplify this as follows

<math>\sum_{k=k_1}^\infty a_{k+r}x^k = \frac{1}{x^r}\sum_{k = k_1}^\infty a_{k + r}x^{k + r} = \frac{1}{x^r} \sum_{l = k_1 + r}^\infty a_lx^l</math>

We now use the ''missing terms in summation'' idea to simplify this final summation in terms of the original.

==Examples==

{| class="sortable" border="1"
! Series we need to sum !! Type !! Rewrite in terms of known series !! Final answer
|-
| <math>\sum_{k=1}^\infty \frac{x^k}{k!}</math> || Missing terms only || <math>\left(\sum_{k=0}^\infty \frac{x^k}{k!} \right) - \frac{x^0}{0!}</math> || <math>e^x - 1</math>
|0
|-
| <math>\sum_{k=0}^\infty \frac{x^k}{(k + 2)!}</math> || Shifted index of summation + missing terms || <math>\frac{1}{x^2}\sum_{l=2}^\infty \frac{x^l}{l!} = \frac{1}{x^2} \left[\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right) - \left(\frac{x^0}{0!} + \frac{x^1}{1!}\right)\right]</math> || <math>\frac{e^x - 1 - x}{x^2}</math>, expression valid for <math>x \ne 0</math>. At <math>x = 0</math>, we get a value of <math>1/2</math>.
|}