Power series - Revision history

Rongzhou: /* Real-world application of geometric series */

2026-09-05T13:35:45Z

Real-world application of geometric series

← Older revision		Revision as of 13:35, 5 September 2026
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	If someone promises to pay you €100 each year until the end of your life on earth, with the first payment made today, how much is that promise worth?		If someone promises to pay you €100 each year until the end of your life on earth, with the first payment made today, how much is that promise worth?

	On first thought, this promise is worth quite a lot of money, especially if you count on living for decades more. However, you must take into account inflation, which in Europe averages out to about 5% per year. The €100 paid today is worth its full €100, a payment of €100 one year from now is worth only €$100/1.05$ in today's money, and a payment in two years is worth only €$(100)/(1.05)^2$ in today's money, and so on. A payment of €100 every year, forever, is therefore worth today		On first thought, this promise is worth quite a lot of money, especially if you count on living for decades more. However, you must take into account inflation, which in Europe averages out to about 5% per year. The €100 paid today is worth its full €100, a payment of €100 one year from now is worth only €$100/1.05$ in today's money, and a payment in two years is worth only €$(100)/(1.05)^2$ in today's money, and so on. A payment of €100 every year, with the first payment made today, forever, is therefore worth today

	$$\sum_{n=0}^{\infty}\frac{100}{1.05^{n}}=2100$$		$$\sum_{0}^{n \to \infty}\frac{100}{1.05^n}= \frac{100}{1-\frac{1}{1.05}}= 2100 $$

	It is still a handsome sum, if you can live on forever.		It is still a handsome sum, if you can live on forever.

RonzzWikiCowriterAI: Annuity example: the first payment is made immediately, so it is worth €100 today and the present-value sum runs from n = 0 and equals €2100. AI-assisted (RonzzWikiCowriter). (via update-page on MediaWiki MCP Server)

2026-09-05T13:31:13Z

Annuity example: the first payment is made immediately, so it is worth €100 today and the present-value sum runs from n = 0 and equals €2100. AI-assisted (RonzzWikiCowriter). (via update-page on MediaWiki MCP Server)

← Older revision		Revision as of 13:31, 5 September 2026
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	=== Real-world application of geometric series ===		=== Real-world application of geometric series ===

	If someone promises to pay you €100 each year until the end of your life on earth, how much is that promise worth?		If someone promises to pay you €100 each year until the end of your life on earth, with the first payment made today, how much is that promise worth?

	On first thought, this promise is worth quite a lot of money, especially if you count on living for decades more. However, you must take into account inflation, which in Europe averages out to about 5% per year. A payment of €100 one year from now is worth only €$100/1.05$ in today's money, and a payment in two years is worth only €$(100)/(1.05)^2$ in today's money, and so on. A payment of €100 every year, forever, is therefore worth today		On first thought, this promise is worth quite a lot of money, especially if you count on living for decades more. However, you must take into account inflation, which in Europe averages out to about 5% per year. The €100 paid today is worth its full €100, a payment of €100 one year from now is worth only €$100/1.05$ in today's money, and a payment in two years is worth only €$(100)/(1.05)^2$ in today's money, and so on. A payment of €100 every year, forever, is therefore worth today

	$$\sum_{n=1}^{\infty}\frac{100}{1.05^{n}}=~~2000~~$$		$$\sum_{n=0}^{\infty}\frac{100}{1.05^{n}}=2100$$

	It is still a handsome sum, if you can live on forever.		It is still a handsome sum, if you can live on forever.

Rongzhou: /* The binomial series */

2026-09-05T13:28:55Z

The binomial series

← Older revision		Revision as of 13:28, 5 September 2026
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	{{#content:Q1689}}		{{#content:Q1689}}

	~~Two cases behave differently and should not be confused:~~

	* '''$p$ a non-negative integer:''' the product $p(p-1)\cdots(p-k+1)$ vanishes once $k$ exceeds $p$, so the series ends after the term in $x^p$ and reduces to the ordinary binomial theorem, which holds for every $x$;
	* '''every other $p$:''' the series continues forever and converges for $-1 < x < 1$ as indicated above; at the endpoints $x = \pm 1$ convergence depends on the value of $p$.

	== Power series solutions of differential equations ==		== Power series solutions of differential equations ==

RonzzWikiCowriterAI: Proofread: fix grammar ("Redivide", stray commas, "there by", "should converges"), correct the derivative formula (multiply by n!/(n-k)!), fix the perpetuity sum (n from 1, value 2000), fix |x|>1 case to x<−1, complete the dangling binomial "two cases", and repair broken math delimiters in the Airy conclusion. AI-assisted (RonzzWikiCowriter). (via update-page on MediaWiki MCP Server)

2026-09-05T13:23:27Z

Proofread: fix grammar ("Redivide", stray commas, "there by", "should converges"), correct the derivative formula (multiply by n!/(n-k)!), fix the perpetuity sum (n from 1, value 2000), fix |x|>1 case to x<−1, complete the dangling binomial "two cases", and repair broken math delimiters in the Airy conclusion. AI-assisted (RonzzWikiCowriter). (via update-page on MediaWiki MCP Server)

← Older revision		Revision as of 13:23, 5 September 2026
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	$$(1-x)S_{N}=(1-x)(1+x+x^{2}+\cdots+x^{N})=(1+x+x^{2}+\cdots+x^{N})-(x+x^{2}+\cdots+x^{N+1})=1-x^{N+1}$$		$$(1-x)S_{N}=(1-x)(1+x+x^{2}+\cdots+x^{N})=(1+x+x^{2}+\cdots+x^{N})-(x+x^{2}+\cdots+x^{N+1})=1-x^{N+1}$$

	~~Redivide~~ both sides by $(1 - x)$; we have ~~therefore~~		Dividing both sides again by $(1 - x)$, we therefore have

	$$S_{N}=\frac{1-x^{N+1}}{1-x}\qquad(x\neq 1)$$ (1)		$$S_{N}=\frac{1-x^{N+1}}{1-x}\qquad(x\neq 1)$$ (1)
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	With $x = 0.1$, for instance, the partial sums run 1; 1.1; 1.11; 1.111; …, settling on 1.111…, and indeed $1/(1 - 0.1) = 1/0.9 = 10/9 = 1.111\ldots$		With $x = 0.1$, for instance, the partial sums run 1; 1.1; 1.11; 1.111; …, settling on 1.111…, and indeed $1/(1 - 0.1) = 1/0.9 = 10/9 = 1.111\ldots$

	* '''When $x = 1$: the series diverges.''' Notice (1) breaks down here, because $1 - x = 0$ and we cannot divide by 0. The sum is however easy to calculate: every term equals 1, so $\lim_{N \to \infty} S_N = \lim_{N \to \infty} N + 1 = \infty$		* '''When $x = 1$: the series diverges.''' Notice (1) breaks down here, because $1 - x = 0$ and we cannot divide by 0. The sum is, however, easy to calculate: every term equals 1, so $\lim_{N \to \infty} S_N = \lim_{N \to \infty} (N + 1) = \infty$

	* '''Case $x = -1$: the series diverges by oscillation.''' The partial sums run 1, 0, 1, 0, 1, … and never settle on a single number. The formula $1/(1 - x)$ would give $1/2$ at $x = -1$, but notice that is the average of 1 and 0. The formula (1) does not guarantee convergence.		* '''Case $x = -1$: the series diverges by oscillation.''' The partial sums run 1, 0, 1, 0, 1, … and never settle on a single number. The formula $1/(1 - x)$ would give $1/2$ at $x = -1$, but notice that is the average of 1 and 0. The formula (1) does not guarantee convergence.

	* '''Case $\|x\| > 1$: the series diverges, as expected.''' When $x >1$, the infinite sum is the sum of an infinite number of increasingly large positive numbers, which goes to infinity. When $x < 1$, the infinite sum oscillates between positive and negative infinity, as it is dominated by the last term, which may be positive or negative.		* '''Case $\|x\| > 1$: the series diverges, as expected.''' When $x > 1$, the infinite sum is the sum of an infinite number of increasingly large positive numbers, which goes to infinity. When $x < -1$, the infinite sum oscillates between positive and negative infinity, as it is dominated by the last term, which may be positive or negative.

	=== Real-world application of geometric series ===		=== Real-world application of geometric series ===

	If someone promises to pay you €100 each year until the end of your life on earth, how much that promise is worth?		If someone promises to pay you €100 each year until the end of your life on earth, how much is that promise worth?

	On first thought, this promise is worth quite a lot of money, especially if you count on living for decades more. However, you must take into account inflation, which in Europe, averages out to about 5% per year. A payment of €100 one year from now is worth only €$100/1.05$ in today's money, and a payment in two years is worth only €$(100)/(1.05)^2$ in today's money, and so on. A payment of €100 every year, forever, is therefore worth today		On first thought, this promise is worth quite a lot of money, especially if you count on living for decades more. However, you must take into account inflation, which in Europe averages out to about 5% per year. A payment of €100 one year from now is worth only €$100/1.05$ in today's money, and a payment in two years is worth only €$(100)/(1.05)^2$ in today's money, and so on. A payment of €100 every year, forever, is therefore worth today

	$$\sum_{0}^{~~n \to~~ \infty}\frac{100}{1.05^n}= ~~2100~~ $$		$$\sum_{n=1}^{\infty}\frac{100}{1.05^{n}}=2000$$

	It is still a handsome sum, if you can live on forever.		It is still a handsome sum, if you can live on forever.
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	== Radius of convergence of a power series ==		== Radius of convergence of a power series ==

	In the previous, section, we have seen that even the simplest power series, a geometric series, may or may not converge depending on the value of $x$. In general, for a power series centred at $c$, there is a number $R \geq 0$, called the '''radius of convergence''', such that:		In the previous section, we have seen that even the simplest power series, a geometric series, may or may not converge depending on the value of $x$. In general, for a power series centred at $c$, there is a number $R \geq 0$, called the '''radius of convergence''', such that:

	* the series converges for every $x$ with $\|x - c\| < R$ (closer to the centre than $R$);		* the series converges for every $x$ with $\|x - c\| < R$ (closer to the centre than $R$);
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	$$f(x)=\sum_{n=0}^{\infty}a_{n}(x-c)^{n}$$		$$f(x)=\sum_{n=0}^{\infty}a_{n}(x-c)^{n}$$

	~~Where~~ $c$ and $a_n$ are unknown.		where $c$ and $a_n$ are unknown.

	Taking the ~~nth~~ derivative of $f(x)$, we have		Taking the $k$-th derivative of $f(x)$, we have

	$$f^{(k)}(x)=\sum_{n=k}^{\infty}\frac~~{a_~~{n}(~~x-c)^{~~n-k}}{\~~frac~~{n!}{(n-k~~)!}~~}$$		$$f^{(k)}(x)=\sum_{n=k}^{\infty}\frac{n!}{(n-k)!}\,a_{n}\,(x-c)^{n-k}$$

	Evaluating at $x=c$, we have		Evaluating at $x=c$, we have
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	$$a_{k}=\frac{f^{(k)}(c)}{k!}$$		$$a_{k}=\frac{f^{(k)}(c)}{k!}$$

	~~Generalise~~ to ~~all possible~~ $n$		Generalising to every $n$, we obtain

	{{#content:Q1688}} (2)		{{#content:Q1688}} (2)
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	{{#content:Q1684}}		{{#content:Q1684}}

	Intuitively, the factorial $n! = 1\cdot 2\cdot 3\cdot\cdots\cdot n$ in the denominator grows much faster than any power function towards infinity, so the terms shrink and the series should ~~converges~~ for every $x$.		Intuitively, the factorial $n! = 1\cdot 2\cdot 3\cdot\cdots\cdot n$ in the denominator grows much faster than any power function towards infinity, so the terms shrink and the series should converge for every $x$.

	Applying the ratio test confirms $R = \infty$:		Applying the ratio test confirms $R = \infty$:
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	$$\lim_{k \to \infty} \frac{\frac{x^{2k+1}}{(2k+1)!}}{\frac{x^{2k+3}}{(2k+3)!}}=\lim_{k \to \infty}\frac{(2k+2)(2k+3)}{x^{2}}= \infty$$		$$\lim_{k \to \infty} \frac{\frac{x^{2k+1}}{(2k+1)!}}{\frac{x^{2k+3}}{(2k+3)!}}=\lim_{k \to \infty}\frac{(2k+2)(2k+3)}{x^{2}}= \infty$$

	~~Which~~ means ~~again,~~ the series converges as expected: to $sin(x)$.		which means the series converges as expected, to $\sin x$.

	== The binomial series ==		== The binomial series ==
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	Two cases behave differently and should not be confused:		Two cases behave differently and should not be confused:

			* '''$p$ a non-negative integer:''' the product $p(p-1)\cdots(p-k+1)$ vanishes once $k$ exceeds $p$, so the series ends after the term in $x^p$ and reduces to the ordinary binomial theorem, which holds for every $x$;
			* '''every other $p$:''' the series continues forever and converges for $-1 < x < 1$ as indicated above; at the endpoints $x = \pm 1$ convergence depends on the value of $p$.

	== Power series solutions of differential equations ==		== Power series solutions of differential equations ==
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	$$\quad\Rightarrow\quad y^{\prime\prime}=C\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^{n},\qquad xy=C\sum_{n=1}^{\infty}a_{n-1}\,x^{n}$$		$$\quad\Rightarrow\quad y^{\prime\prime}=C\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^{n},\qquad xy=C\sum_{n=1}^{\infty}a_{n-1}\,x^{n}$$

	Substituting these two expressions back into (3) and collecting terms, noting the equality must hold for every value of $x$		Substituting these two expressions back into (3) and collecting terms, noting that the equality must hold for every value of $x$, gives

	$$\quad a_{2}=0,\qquad (n+2)(n+1)\,a_{n+2}=a_{n-1}\qquad(n\geq 1)$$		$$\quad a_{2}=0,\qquad (n+2)(n+1)\,a_{n+2}=a_{n-1}\qquad(n\geq 1)$$

	We can ~~there by~~ express all the coefficients in terms of $a_0$ and $a_1$:		We can thereby express all the coefficients in terms of $a_0$ and $a_1$:

	$$a_3 = a_0/(3\cdot 2) = a_0/6, \qquad a_6 = a_3/(6\cdot 5) = a_0/180,$$		$$a_3 = a_0/(3\cdot 2) = a_0/6, \qquad a_6 = a_3/(6\cdot 5) = a_0/180,$$
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	$$y=C (a_{0}\left(1+\frac{x^{3}}{6}+\frac{x^{6}}{180}+\cdots\right)+a_{1}\left(x+\frac{x^{4}}{12}+\frac{x^{7}}{504}+\cdots\right))$$		$$y=C (a_{0}\left(1+\frac{x^{3}}{6}+\frac{x^{6}}{180}+\cdots\right)+a_{1}\left(x+\frac{x^{4}}{12}+\frac{x^{7}}{504}+\cdots\right))$$

	The ratio test gives $R = \infty$~~. Therefore~~ each series converges for every $x$. Setting $C_1=Ca_0$, C_2=Ca_1$, and $~~A_i=1+\frac{~~x~~^{3}}{6}+\frac{~~x~~^{6}}{180}+\cdot~~, ~~B_i=x+\frac{x^{4}}{12}+\frac{x^{7}}{504}+\cdots$~~ we arrive at the well-known form		The ratio test gives $R = \infty$, so each series converges for every $x$. Setting $C_1 = Ca_0$ and $C_2 = Ca_1$, and denoting the two bracketed series by $A(x)$ and $B(x)$, we arrive at the well-known form

	{{#content:Q1609}}		{{#content:Q1609}}

Rongzhou at 13:11, 5 September 2026

2026-09-05T13:11:04Z

Show changes

SeedBot: quote-guard deployed (SimpleMathJax internal-parse patch): revert the y^{\prime\prime} workaround to natural TeX primes $y''$ — '' inside math now survives wikitext emphasis parsing. AI-assisted (RonzzWikiCowriter).

2026-09-05T12:38:46Z

quote-guard deployed (SimpleMathJax internal-parse patch): revert the y^{\prime\prime} workaround to natural TeX primes $y''$ — '' inside math now survives wikitext emphasis parsing. AI-assisted (RonzzWikiCowriter).

← Older revision		Revision as of 12:38, 5 September 2026
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	The Airy equation		The Airy equation

	$$y~~^{\prime\prime}~~ - xy = 0$$		$$y'' - xy = 0$$

	is named after [[Person:George Biddell Airy\|George Biddell Airy]], who met it in 1838 while studying the intensity of light near a caustic. Its solutions cannot be written as finite combinations of elementary functions, so it is the standard test case for series methods.		is named after [[Person:George Biddell Airy\|George Biddell Airy]], who met it in 1838 while studying the intensity of light near a caustic. Its solutions cannot be written as finite combinations of elementary functions, so it is the standard test case for series methods.

	Near $x = 0$ the coefficients of the equation are as well behaved as possible (the point is an ordinary point), so seek a solution as a power series about 0: $y = a_0 + a_1 x + a_2 x^2 + \cdots$, where the $a_n$ are unknown constants. Differentiate term by term: $y~~^{\prime}~~ = a_1 + 2a_2 x + 3a_3 x^2 + \cdots$ and $y~~^{\prime\prime}~~ = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots$, so each derivative of $x^n$ lowers the power by one and multiplies the coefficient by $n$. Re-indexing the sums so that both run over the same powers of $x$ gives		Near $x = 0$ the coefficients of the equation are as well behaved as possible (the point is an ordinary point), so seek a solution as a power series about 0: $y = a_0 + a_1 x + a_2 x^2 + \cdots$, where the $a_n$ are unknown constants. Differentiate term by term: $y' = a_1 + 2a_2 x + 3a_3 x^2 + \cdots$ and $y'' = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots$, so each derivative of $x^n$ lowers the power by one and multiplies the coefficient by $n$. Re-indexing the sums so that both run over the same powers of $x$ gives

	$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad y~~^{\prime\prime}~~=\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^{n},\qquad xy=\sum_{n=1}^{\infty}a_{n-1}\,x^{n}$$		$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad y''=\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^{n},\qquad xy=\sum_{n=1}^{\infty}a_{n-1}\,x^{n}$$

	Substituting these two expressions into $y~~^{\prime\prime}~~ - xy = 0$ and collecting terms, the equality must hold for every value of $x$; that forces the coefficient of each power of $x$ on the left to vanish. The constant term gives $2\cdot 1\cdot a_2 = 0$, so $a_2 = 0$, and comparing the coefficient of $x^n$ for $n \geq 1$ gives the recurrence		Substituting these two expressions into $y'' - xy = 0$ and collecting terms, the equality must hold for every value of $x$; that forces the coefficient of each power of $x$ on the left to vanish. The constant term gives $2\cdot 1\cdot a_2 = 0$, so $a_2 = 0$, and comparing the coefficient of $x^n$ for $n \geq 1$ gives the recurrence

	$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad a_{2}=0,\qquad (n+2)(n+1)\,a_{n+2}=a_{n-1}\qquad(n\geq 1)$$		$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad a_{2}=0,\qquad (n+2)(n+1)\,a_{n+2}=a_{n-1}\qquad(n\geq 1)$$
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	=== Case 2: expanding about a singular point, the Frobenius method ===		=== Case 2: expanding about a singular point, the Frobenius method ===

	Write a linear second-order equation in the form $y~~^{\prime\prime}~~ + p(x)y~~^{\prime}~~ + q(x)y = 0$. A point $x_0$ is ordinary when both $p$ and $q$ can be expanded in power series about $x_0$; the direct method of Case 1 applies there. If at least one of $p$ or $q$ cannot, typically because it divides by $x - x_0$, the point is singular. The point is a regular singular point when $(x - x_0)\,p(x)$ and $(x - x_0)^2 q(x)$ do have power series expansions about $x_0$.		Write a linear second-order equation in the form $y'' + p(x)y' + q(x)y = 0$. A point $x_0$ is ordinary when both $p$ and $q$ can be expanded in power series about $x_0$; the direct method of Case 1 applies there. If at least one of $p$ or $q$ cannot, typically because it divides by $x - x_0$, the point is singular. The point is a regular singular point when $(x - x_0)\,p(x)$ and $(x - x_0)^2 q(x)$ do have power series expansions about $x_0$.

	At a singular point a plain power series cannot work: with $p$ or $q$ dividing by $x$, the equation mixes powers whose lowest terms start at different heights, so they cannot be matched. The Frobenius method instead seeks a solution with a shifted leading power, $y = x^r (a_0 + a_1 x + \cdots)$. Substituting and setting the coefficient of the lowest power of $x$ to zero produces an algebraic equation, the indicial equation, whose solutions fix the possible exponents $r$. For the Bessel equation of order $\nu$,		At a singular point a plain power series cannot work: with $p$ or $q$ dividing by $x$, the equation mixes powers whose lowest terms start at different heights, so they cannot be matched. The Frobenius method instead seeks a solution with a shifted leading power, $y = x^r (a_0 + a_1 x + \cdots)$. Substituting and setting the coefficient of the lowest power of $x$ to zero produces an algebraic equation, the indicial equation, whose solutions fix the possible exponents $r$. For the Bessel equation of order $\nu$,

SeedBot: fix mangled derivative math in the differential-equations section: MathJax treats '' inside $y''$/$y'$ delimiters as a closing quote and terminates the math (dangling $ then swallowed prose). Write derivatives as $y^{\prime}$ / $y^{\prime\prime}$. AI-assisted (RonzzWikiCowriter).

2026-09-04T15:57:29Z

fix mangled derivative math in the differential-equations section: MathJax treats '' inside $y''$/$y'$ delimiters as a closing quote and terminates the math (dangling $ then swallowed prose). Write derivatives as $y^{\prime}$ / $y^{\prime\prime}$. AI-assisted (RonzzWikiCowriter).

← Older revision		Revision as of 15:57, 4 September 2026
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	The Airy equation		The Airy equation

	$$y'' - xy = 0$$		$$y^{\prime\prime} - xy = 0$$

	is named after [[Person:George Biddell Airy\|George Biddell Airy]], who met it in 1838 while studying the intensity of light near a caustic. Its solutions cannot be written as finite combinations of elementary functions, so it is the standard test case for series methods.		is named after [[Person:George Biddell Airy\|George Biddell Airy]], who met it in 1838 while studying the intensity of light near a caustic. Its solutions cannot be written as finite combinations of elementary functions, so it is the standard test case for series methods.

	Near $x = 0$ the coefficients of the equation are as well behaved as possible (the point is an ordinary point), so seek a solution as a power series about 0: $y = a_0 + a_1 x + a_2 x^2 + \cdots$, where the $a_n$ are unknown constants. Differentiate term by term: $y' = a_1 + 2a_2 x + 3a_3 x^2 + \cdots$ and $y'' = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots$, so each derivative of $x^n$ lowers the power by one and multiplies the coefficient by $n$. Re-indexing the sums so that both run over the same powers of $x$ gives		Near $x = 0$ the coefficients of the equation are as well behaved as possible (the point is an ordinary point), so seek a solution as a power series about 0: $y = a_0 + a_1 x + a_2 x^2 + \cdots$, where the $a_n$ are unknown constants. Differentiate term by term: $y^{\prime} = a_1 + 2a_2 x + 3a_3 x^2 + \cdots$ and $y^{\prime\prime} = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots$, so each derivative of $x^n$ lowers the power by one and multiplies the coefficient by $n$. Re-indexing the sums so that both run over the same powers of $x$ gives

	$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad y''=\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^{n},\qquad xy=\sum_{n=1}^{\infty}a_{n-1}\,x^{n}$$		$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad y^{\prime\prime}=\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^{n},\qquad xy=\sum_{n=1}^{\infty}a_{n-1}\,x^{n}$$

	Substituting these two expressions into $y'' - xy = 0$ and collecting terms, the equality must hold for every value of $x$; that forces the coefficient of each power of $x$ on the left to vanish. The constant term gives $2\cdot 1\cdot a_2 = 0$, so $a_2 = 0$, and comparing the coefficient of $x^n$ for $n \geq 1$ gives the recurrence		Substituting these two expressions into $y^{\prime\prime} - xy = 0$ and collecting terms, the equality must hold for every value of $x$; that forces the coefficient of each power of $x$ on the left to vanish. The constant term gives $2\cdot 1\cdot a_2 = 0$, so $a_2 = 0$, and comparing the coefficient of $x^n$ for $n \geq 1$ gives the recurrence

	$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad a_{2}=0,\qquad (n+2)(n+1)\,a_{n+2}=a_{n-1}\qquad(n\geq 1)$$		$$y=\sum_{n=0}^{\infty}a_{n}x^{n}\quad\Rightarrow\quad a_{2}=0,\qquad (n+2)(n+1)\,a_{n+2}=a_{n-1}\qquad(n\geq 1)$$
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	=== Case 2: expanding about a singular point, the Frobenius method ===		=== Case 2: expanding about a singular point, the Frobenius method ===

	Write a linear second-order equation in the form $y'' + p(x)y' + q(x)y = 0$. A point $x_0$ is ordinary when both $p$ and $q$ can be expanded in power series about $x_0$; the direct method of Case 1 applies there. If at least one of $p$ or $q$ cannot, typically because it divides by $x - x_0$, the point is singular. The point is a regular singular point when $(x - x_0)\,p(x)$ and $(x - x_0)^2 q(x)$ do have power series expansions about $x_0$.		Write a linear second-order equation in the form $y^{\prime\prime} + p(x)y^{\prime} + q(x)y = 0$. A point $x_0$ is ordinary when both $p$ and $q$ can be expanded in power series about $x_0$; the direct method of Case 1 applies there. If at least one of $p$ or $q$ cannot, typically because it divides by $x - x_0$, the point is singular. The point is a regular singular point when $(x - x_0)\,p(x)$ and $(x - x_0)^2 q(x)$ do have power series expansions about $x_0$.

	At a singular point a plain power series cannot work: with $p$ or $q$ dividing by $x$, the equation mixes powers whose lowest terms start at different heights, so they cannot be matched. The Frobenius method instead seeks a solution with a shifted leading power, $y = x^r (a_0 + a_1 x + \cdots)$. Substituting and setting the coefficient of the lowest power of $x$ to zero produces an algebraic equation, the indicial equation, whose solutions fix the possible exponents $r$. For the Bessel equation of order $\nu$,		At a singular point a plain power series cannot work: with $p$ or $q$ dividing by $x$, the equation mixes powers whose lowest terms start at different heights, so they cannot be matched. The Frobenius method instead seeks a solution with a shifted leading power, $y = x^r (a_0 + a_1 x + \cdots)$. Substituting and setting the coefficient of the lowest power of $x$ to zero produces an algebraic equation, the indicial equation, whose solutions fix the possible exponents $r$. For the Bessel equation of order $\nu$,

SeedBot: math convention: in-page $…$ inline / $$…$$ blocks for demonstrations + numeric examples; {{#content:}} kept only for the named/known formulas (general form Q1667, coefficient/Taylor Q1688, e^x Q1684, sin x Q1686, binomial Q1689, Airy Ai/Bi Q1609, Bessel Q1670); raw ''a''₀-style markup converted to inline math. AI-assisted (RonzzWikiCowriter). (via update-page on MediaWiki MCP Server)

2026-09-04T15:52:48Z

math convention: in-page $…$ inline / $$…$$ blocks for demonstrations + numeric examples; {{#content:}} kept only for the named/known formulas (general form Q1667, coefficient/Taylor Q1688, e^x Q1684, sin x Q1686, binomial Q1689, Airy Ai/Bi Q1609, Bessel Q1670); raw ''a''₀-style markup converted to inline math. AI-assisted (RonzzWikiCowriter). (via update-page on MediaWiki MCP Server)

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Rongzhou at 15:38, 4 September 2026

2026-09-04T15:38:51Z

← Older revision		Revision as of 15:38, 4 September 2026
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	'''A power series''' is an infinite sum ~~built from~~ the ~~powers of a variable: a sum with infinitely many terms, each a constant times a power of ''x''. The~~ general form~~, centred at a number ''c'', is~~		'''A power series''' is an infinite sum with the general form

	{{#content:Q1667}}		{{#content:Q1667}}

	The numbers ''a''₀, ''a''₁, ''a''₂, … are the '''coefficients''', and ''c'' ~~is the '''centre''' of the series. When ''c'' = 0~~, ~~the powers are simply ''x''<sup>''n''</sup> and the series reads ''~~a~~''₀ + ''a''₁''x'' + ''a''₂''x''² + ⋯.~~		The numbers ''a''₀, ''a''₁, ''a''₂, … are constants, known as the '''coefficients''', and ''c'', also a constant, is known as the '''centre''' of the series.

	~~A power series~~ is~~, in plain words, a polynomial with infinitely many terms. Its usefulness comes from~~ the ~~fact that its terms require only the four basic operations of arithmetic to evaluate: each term '~~'a''~~<sub>''n''</sub>''x''<sup>'~~'n''~~</sup> is a multiplication followed by an addition. Stopping after finitely many terms leaves a polynomial that approximates the value~~ of the whole sum, and keeping more terms improves the approximation (wherever the series converges). Numbers that have no formula built from basic operations alone, such as e<sup>0.05</sup> or sin 1, can therefore be computed to any desired accuracy by arithmetic only. This is the idea behind the scientific functions on calculators and in spreadsheets, and it is the reason power series ~~pervade science~~.

	== The geometric series ==		== The geometric series ==
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	:1 + ''x'' + ''x''² + ''x''³ + ⋯		:1 + ''x'' + ''x''² + ''x''³ + ⋯

	~~Each~~ term multiplies the previous one by ''x''~~; this~~ is ~~called~~ the '''geometric series'''. ~~It is~~ the ~~right place to start, because its~~ sum ~~can be worked out exactly~~, ~~and because it shows clearly when an infinite sum produces a number and when it does not.~~		As seen, each term multiplies the previous one by ''x''. This is known as the '''geometric series'''.

			For a geometric series, we can calculate the partial sum of the first ''N'' + 1 terms, ''S''<sub>''N''</sub> = 1 + ''x'' + ''x''² + ⋯ + ''x''<sup>''N''</sup>:

	~~Start with the finite version, the partial sum of the first ''N'' + 1 terms, ''S''<sub>''N''</sub> = 1 + ''x'' + ''x''² + ⋯ + ''x''<sup>''N''</sup>.~~ Multiply both sides of this equation by (1 − ''x'') and ~~multiply out~~ the brackets:		Multiply both sides of this equation by (1 − ''x'') and eliminate the brackets:

	{{#content:Q1681}}		{{#content:Q1681}}

	~~Every middle term cancels: the two brackets share the terms ''x'', ''x''², …, ''x''<sup>''N''</sup>, which subtract to nothing and leave only 1 − ''x''<sup>''N''+1</sup>. Dividing~~ both sides by (1 − ''x''), ~~which is allowed for any ''x'' ≠ 1, gives the closed form of the partial sum:~~		Redivide both sides by (1 − ''x''), we have therefore

	{{#content:Q1682}}		{{#content:Q1682}}

RonzzWikiCowriterAI: Rewrite for an undergraduate, non-math-specialist audience: build-up from the geometric series, case-by-case convergence, general case + worked numeric examples, full derivations; series solutions of ODEs kept as the advanced application. AI-assisted (RonzzWikiCowriter) (via update-page on MediaWiki MCP Server)

2026-09-04T15:21:47Z

Rewrite for an undergraduate, non-math-specialist audience: build-up from the geometric series, case-by-case convergence, general case + worked numeric examples, full derivations; series solutions of ODEs kept as the advanced application. AI-assisted (RonzzWikiCowriter) (via update-page on MediaWiki MCP Server)

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