Substitution method for power series summation - Revision history

Vipul: /* Interval of validity of power series summation */

2012-07-06T23:10:57Z

Interval of validity of power series summation

← Older revision		Revision as of 23:10, 6 July 2012
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	In the good case that the formula is valid over the entire interval of convergence for the power series in <math>u</math>, we have:		In the good case that the formula is valid over the entire interval of convergence for the power series in <math>u</math>, we have:

			* If the power series in <math>u</math> converges everywhere to the function we have obtained, then the power series in <math>x</math> also converges everywhere to the function we have obtained.
	* If the radius of convergence for the power series in <math>u</math> is <math>R</math>, the radius of convergence of the power series in <math>x</math> is <math>(R/\|\lambda\|)^{1/m}</math>.		* If the radius of convergence for the power series in <math>u</math> is <math>R</math>, the radius of convergence of the power series in <math>x</math> is <math>(R/\|\lambda\|)^{1/m}</math>.
	* To determine whether the endpoints are included, apply the map <math>x \mapsto \lambda x^m</math> to the values <math>x = \pm (R/\|\lambda\|)^{1/m}</math> and check whether the image of the map is in the interval of convergence for <math>u</math>.		* To determine whether the endpoints are included, apply the map <math>x \mapsto \lambda x^m</math> to the values <math>x = \pm (R/\|\lambda\|)^{1/m}</math> and check whether the image of the map is in the interval of convergence for <math>u</math>.

	~~Note~~ in ~~particular that~~ if the ~~power series~~ in <math>u</math> ~~converges everywhere to~~ the ~~function we have obtained~~, ~~then~~ the ~~power series~~ in <math>x</math> ~~also converges everywhere to~~ the ~~function we have obtained~~.		Let's now consider some of the finite cases more explicitly:

			{\| class="sortable" border="1"
			! Sign of <math>\lambda</math> !! Parity of <math>m</math> !! What can we say?
			\|-
			\| positive \|\| odd \|\| The negative endpoint <math>-(R/\|\lambda\|)^{1/m}</math> is in the interval of convergence for <math>x</math> if and only if <math>-R</math> is in the interval of convergence for <math>u</math>, and the positive endpoint <math>(R/\|\lambda\|)^{1/m}</math> is in the interval of convergence for <math>x</math> if and only if <math>R</math> is in the interval of convergence for <math>u</math>.
			\|-
			\| positive \|\| even \|\| Both endpoints <math>\pm (R/\|\lambda\|)^{1/m}</math> are in the interval of convergence for <math>x</math> if and only if <math>R</math> is in the interval of convergence for <math>u</math>. Otherwise, neither endpoint is included.
			\|-
			\| negative \|\| odd \|\| The negative endpoint <math>-(R/\|\lambda\|)^{1/m}</math> is in the interval of convergence for <math>x</math> if and only if <math>R</math> is in the interval of convergence for <math>u</math>, and the positive endpoint <math>(R/\|\lambda\|)^{1/m}</math> is in the interval of convergence for <math>x</math> if and only if <math>-R</math> is in the interval of convergence for <math>u</math>.
			\|-
			\| negative \|\| even \|\| Both endpoints <math>\pm (R/\|\lambda\|)^{1/m}</math> are in the interval of convergence for <math>x</math> if and only if <math>-R</math> is in the interval of convergence for <math>u</math>. Otherwise, neither endpoint is included.
			\|}
	===Multiplication and division===		===Multiplication and division===

	If the substitution method is combined with multiplying/dividing by a power of <math>x</math>, then we also need to worry about the case <math>x = 0</math>. In this case, the power series may still converge, but the formal expression we obtained may not be valid, though its limit would still be the right value.		If the substitution method is combined with multiplying/dividing by a power of <math>x</math>, then we also need to worry about the case <math>x = 0</math>. In this case, the power series may still converge, but the formal expression we obtained may not be valid, though its limit would still be the right value.

Vipul: /* Interval of validity of power series summation */

2012-07-06T22:53:13Z

Interval of validity of power series summation

← Older revision		Revision as of 22:53, 6 July 2012
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	* To determine whether the endpoints are included, apply the map <math>x \mapsto \lambda x^m</math> to the values <math>x = \pm (R/\|\lambda\|)^{1/m}</math> and check whether the image of the map is in the interval of convergence for <math>u</math>.		* To determine whether the endpoints are included, apply the map <math>x \mapsto \lambda x^m</math> to the values <math>x = \pm (R/\|\lambda\|)^{1/m}</math> and check whether the image of the map is in the interval of convergence for <math>u</math>.

			Note in particular that if the power series in <math>u</math> converges everywhere to the function we have obtained, then the power series in <math>x</math> also converges everywhere to the function we have obtained.
	===Multiplication and division===		===Multiplication and division===

	If the substitution method is combined with multiplying/dividing by a power of <math>x</math>, then we also need to worry about the case <math>x = 0</math>. In this case, the power series may still converge, but the formal expression we obtained may not be valid, though its limit would still be the right value.		If the substitution method is combined with multiplying/dividing by a power of <math>x</math>, then we also need to worry about the case <math>x = 0</math>. In this case, the power series may still converge, but the formal expression we obtained may not be valid, though its limit would still be the right value.

Vipul: /* Application */

2012-07-06T22:31:10Z

Application

← Older revision		Revision as of 22:31, 6 July 2012
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	Substitution can be combined with another common technique for power series manipulation: multiply and divide by <math>x</math> in order to make the exponent better match the coefficient.		Substitution can be combined with another common technique for power series manipulation: multiply and divide by <math>x</math> in order to make the exponent better match the coefficient.

			==Related methods==

			* [[Exponent shift method for power series summation]]
			* [[Integration and differentiation method for power series summation]]

	==Examples==		==Examples==

Vipul: /* Interaction with interval of convergence */

2012-07-06T22:28:19Z

Interaction with interval of convergence

← Older revision		Revision as of 22:28, 6 July 2012
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	<math>\sum_{k=0}^\infty \frac{x^k}{(2k)!} = \left\lbrace \begin{array}{rl} \cosh \sqrt{x}, & x \ge 0 \\ \cos \sqrt{-x}, & x < 0 \\\end{array}\right.</math>		<math>\sum_{k=0}^\infty \frac{x^k}{(2k)!} = \left\lbrace \begin{array}{rl} \cosh \sqrt{x}, & x \ge 0 \\ \cos \sqrt{-x}, & x < 0 \\\end{array}\right.</math>

	==~~Interaction with interval~~ of ~~convergence~~==		==Interval of validity of power series summation==

	If the ~~power series after~~ the ~~<math>u</math>-~~substitution ~~<math>u~~ = ~~\lambda x^m</math> has a radius of convergence <math>R</math> for <math>u</math>, the original power series has a radius of convergence <math>(R/\|\lambda\|)^{1/m}</math>.~~		===Transforming the interval of validity along with the substitution===

	~~For inclusion~~ of ~~endpoints~~, we ~~need to be more careful~~. ~~Essentially~~, we ~~need to figure out how~~ the ~~substitution~~ <math>u = \lambda x^m</math> ~~behaves on~~ <math>x = \pm (R/\|\lambda\|)^{1/m}</math>.		Consider the <math>u</math>-substitution <math>u = \lambda x^m</math>, giving:

			<math>\sum a_k\lambda^kx^{mk} \to \sum a_ku^k</math>

			We use some formula to sum up the power series rewritten in terms of <math>u</math>. In the typical scenario, the formula we use is valid over the entire [[interval of convergence]] of <math>u</math>, though we may sometimes have a situation where the formula we've obtained is valid only on part of the interval. We then plug back <math>u = x^{mk}</math> and get a formula for the summation in terms of <math>x</math>.

			The final formula is valid ''precisely'' for those <math>x</math> for which <math>x^{mk}</math> is in the set of values of <math>u</math> for which the formula in terms of <math>u</math> is valid.

			In the good case that the formula is valid over the entire interval of convergence for the power series in <math>u</math>, we have:

			* If the radius of convergence for the power series in <math>u</math> is <math>R</math>, the radius of convergence of the power series in <math>x</math> is <math>(R/\|\lambda\|)^{1/m}</math>.
			* To determine whether the endpoints are included, apply the map <math>x \mapsto \lambda x^m</math> to the values <math>x = \pm (R/\|\lambda\|)^{1/m}</math> and check whether the image of the map is in the interval of convergence for <math>u</math>.

			===Multiplication and division===

			If the substitution method is combined with multiplying/dividing by a power of <math>x</math>, then we also need to worry about the case <math>x = 0</math>. In this case, the power series may still converge, but the formal expression we obtained may not be valid, though its limit would still be the right value.

Vipul: /* Simple examples */

2012-07-06T22:07:41Z

Simple examples

← Older revision		Revision as of 22:07, 6 July 2012
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	\| <math>\sum_{k=0}^\infty \frac{x^{3k}}{k!}</math> \|\| <math>u = x^3</math> \|\| power \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{x^3}</math>		\| <math>\sum_{k=0}^\infty \frac{x^{3k}}{k!}</math> \|\| <math>u = x^3</math> \|\| power \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{x^3}</math>
	\|-		\|-
	\| <math>\sum_{k=1}^\infty \frac{x^{5k}}{k}</math> \|\| <math>u = x^5</math> \|\| power \|\| <math>\sum_{k=1}^\infty \frac{u^k}{k}</math> \|\| <math>-\ln(1 - u)</math> \|\| <math>-\ln(1 - x^5)</math> (note that the interval of convergence is <math>[-1,1)</math>)		\| <math>\sum_{k=1}^\infty \frac{x^{5k}}{k}</math> \|\| <math>u = x^5</math> \|\| power \|\| <math>\sum_{k=1}^\infty \frac{u^k}{k}</math> \|\| <math>-\ln(1 - u)</math> \|\| <math>-\ln(1 - x^5)</math> (note that the interval of convergence is not all of <math>\R</math>, but rather is <math>[-1,1)</math> -- however, that is not the focus here)
	\|-		\|-
	\| <math>\sum_{k=0}^\infty \frac{2^kx^k}{k!}</math> \|\| <math>u = 2x</math> \|\| scalar multiple \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{2x}</math>		\| <math>\sum_{k=0}^\infty \frac{2^kx^k}{k!}</math> \|\| <math>u = 2x</math> \|\| scalar multiple \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{2x}</math>

Vipul: /* Simple examples */

2012-07-06T22:07:12Z

Simple examples

← Older revision		Revision as of 22:07, 6 July 2012
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	\| <math>\sum_{k=0}^\infty \frac{x^{3k}}{k!}</math> \|\| <math>u = x^3</math> \|\| power \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{x^3}</math>		\| <math>\sum_{k=0}^\infty \frac{x^{3k}}{k!}</math> \|\| <math>u = x^3</math> \|\| power \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{x^3}</math>
	\|-		\|-
	\| <math>\sum_{k=1}^\infty \frac{x^{5k}}{k}</math> \|\| <math>u = x^5</math> \|\| power \|\| <math>\sum_{k=1}^\infty \frac{u^k}{k}</math> \|\| <math>-\ln(1 - u)</math> \|\| <math>-\ln(1 - x^5)</math> (note that the interval of convergence is <math>(-1,1)</math>		\| <math>\sum_{k=1}^\infty \frac{x^{5k}}{k}</math> \|\| <math>u = x^5</math> \|\| power \|\| <math>\sum_{k=1}^\infty \frac{u^k}{k}</math> \|\| <math>-\ln(1 - u)</math> \|\| <math>-\ln(1 - x^5)</math> (note that the interval of convergence is <math>[-1,1)</math>)
	\|-		\|-
	\| <math>\sum_{k=0}^\infty \frac{2^kx^k}{k!}</math> \|\| <math>u = 2x</math> \|\| scalar multiple \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{2x}</math>		\| <math>\sum_{k=0}^\infty \frac{2^kx^k}{k!}</math> \|\| <math>u = 2x</math> \|\| scalar multiple \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{2x}</math>

Vipul: /* Interaction with interval of convergence */

2012-07-06T22:06:30Z

Interaction with interval of convergence

← Older revision		Revision as of 22:06, 6 July 2012
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	If the power series after the <math>u</math>-substitution <math>u = \lambda x^m</math> has a radius of convergence <math>R</math> for <math>u</math>, the original power series has a radius of convergence <math>(R/\|\lambda\|)^{1/m}</math>.		If the power series after the <math>u</math>-substitution <math>u = \lambda x^m</math> has a radius of convergence <math>R</math> for <math>u</math>, the original power series has a radius of convergence <math>(R/\|\lambda\|)^{1/m}</math>.

	For inclusion of endpoints, we need to be more careful. We need to ~~make cases based on~~ the ~~sign of~~ <math>\lambda</math> ~~and whether~~ <math>m</~~math> is even or odd (assuming <math>~~m</math> ~~is an integer)~~.		For inclusion of endpoints, we need to be more careful. Essentially, we need to figure out how the substitution <math>u = \lambda x^m</math> behaves on <math>x = \pm (R/\|\lambda\|)^{1/m}</math>.

Vipul: /* Example involving a fractional power substitution */

2012-07-06T22:03:00Z

Example involving a fractional power substitution

← Older revision		Revision as of 22:03, 6 July 2012
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	<math>\sum_{k=0}^\infty \frac{x^k}{(2k)!} = \left\lbrace \begin{array}{rl} \cosh \sqrt{x}, & x \ge 0 \\ \cos \sqrt{-x}, & x < 0 \\\end{array}\right.</math>		<math>\sum_{k=0}^\infty \frac{x^k}{(2k)!} = \left\lbrace \begin{array}{rl} \cosh \sqrt{x}, & x \ge 0 \\ \cos \sqrt{-x}, & x < 0 \\\end{array}\right.</math>

			==Interaction with interval of convergence==

			If the power series after the <math>u</math>-substitution <math>u = \lambda x^m</math> has a radius of convergence <math>R</math> for <math>u</math>, the original power series has a radius of convergence <math>(R/\|\lambda\|)^{1/m}</math>.

			For inclusion of endpoints, we need to be more careful. We need to make cases based on the sign of <math>\lambda</math> and whether <math>m</math> is even or odd (assuming <math>m</math> is an integer).

Vipul: /* Simple examples */

2012-07-06T21:57:39Z

Simple examples

← Older revision		Revision as of 21:57, 6 July 2012
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	\| <math>\sum_{k=0}^\infty \frac{x^{3k}}{k!}</math> \|\| <math>u = x^3</math> \|\| power \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{x^3}</math>		\| <math>\sum_{k=0}^\infty \frac{x^{3k}}{k!}</math> \|\| <math>u = x^3</math> \|\| power \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{x^3}</math>
	\|-		\|-
	\| <math>\sum_{k=1}^\infty \frac{x^{5k}}{k}</math> \|\| <math>u = x^5</math> \|\| power \|\| <math>\sum_{k=1}^\infty \frac{u^k}{k}</math> \|\| <math>-\ln(1 - u)</math> \|\| <math>-\ln(1 - x^5)</math>		\| <math>\sum_{k=1}^\infty \frac{x^{5k}}{k}</math> \|\| <math>u = x^5</math> \|\| power \|\| <math>\sum_{k=1}^\infty \frac{u^k}{k}</math> \|\| <math>-\ln(1 - u)</math> \|\| <math>-\ln(1 - x^5)</math> (note that the interval of convergence is <math>(-1,1)</math>
	\|-		\|-
	\| <math>\sum_{k=0}^\infty \frac{2^kx^k}{k!}</math> \|\| <math>u = 2x</math> \|\| scalar multiple \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{2x}</math>		\| <math>\sum_{k=0}^\infty \frac{2^kx^k}{k!}</math> \|\| <math>u = 2x</math> \|\| scalar multiple \|\| <math>\sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>e^u</math> \|\| <math>e^{2x}</math>

Vipul: /* Examples that involve some combination of substitution and multiplying/dividing */

2012-07-06T21:47:33Z

Examples that involve some combination of substitution and multiplying/dividing

← Older revision		Revision as of 21:47, 6 July 2012
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	\| <math>\sum_{k=0}^\infty \frac{x^{3k + 1}}{k!}</math> \|\| first pull out a factor of <math>x</math>, then do <math>u = x^3</math> \|\| power \|\| <math>x \sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>xe^u</math> \|\| <math>xe^{x^3}</math>		\| <math>\sum_{k=0}^\infty \frac{x^{3k + 1}}{k!}</math> \|\| first pull out a factor of <math>x</math>, then do <math>u = x^3</math> \|\| power \|\| <math>x \sum_{k=0}^\infty \frac{u^k}{k!}</math> \|\| <math>xe^u</math> \|\| <math>xe^{x^3}</math>
	\|-		\|-
	\| <math>\sum_{k=0}^\infty \frac{x^{14k + 2}}{(2k + 1)!}</math> \|\| first multiply/divide by <math>x^5</math> (to make the exponent <math>7(2k + 1)</math>), then do <math>u = x^7</math> \|\| power \|\| <math>\frac{1}{x^5} \sum_{k=0}^\infty \frac{u^{2k + 1}}{(2k + 1)!}</math> \|\| <math>\frac{\sinh u}{x^5}</math> \|\| <math>\frac{\sinh(x^7)}{x^5}</math>		\| <math>\sum_{k=0}^\infty \frac{x^{14k + 2}}{(2k + 1)!}</math> \|\| first multiply/divide by <math>x^5</math> (to make the exponent <math>7(2k + 1)</math>), then do <math>u = x^7</math> \|\| power \|\| <math>\frac{1}{x^5} \sum_{k=0}^\infty \frac{u^{2k + 1}}{(2k + 1)!}</math> \|\| <math>\frac{\sinh u}{x^5}</math> \|\| <math>\frac{\sinh(x^7)}{x^5}</math> (Note: The expression is not valid at <math>x = 0</math> though its limit equals the power series sum at 0; both the limit and the power series sum at 0 are equal to 0)
	\|}		\|}