Rafael Davis 4:07 am on June 7, 2017
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Cauchy–Euler equation & power series

Cauchy–Euler equation

x^{2}{\frac {d^{2}y}{dx^{2}}}+ax{\frac {dy}{dx}}+by=0.\,

  • Case #1: Two distinct roots, m1 and m2
  • Case #2: One real repeated root, m
  • Case #3: Complex roots, α ± βi
In case #1, the solution is given by:
y=c_{1}x^{{m_{{1}}}}+c_{2}x^{{m_{2}}}\,
In case #2, the solution is given by
y=c_{1}x^{m}\ln(x)+c_{2}x^{m}\,
To get to this solution, the method of reduction of order must be applied after having found one solution y = xm.
In case #3, the solution is given by:
y=c_{1}x^{\alpha }\cos(\beta \ln(x))+c_{2}x^{\alpha }\sin(\beta \ln(x))\,
\alpha ={\mathop {{\rm {Re}}}}(m)\,
\beta ={\mathop {{\rm {Im}}}}(m)\,
For c_{1},c_{2} ∈ ℝ .

List of Maclaurin series of some common functions[edit]

\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}\,(x-a)^{n}

{\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots }.
{\displaystyle {\begin{aligned}{\frac {1}{1-x}}&=\sum _{n=0}^{\infty }x^{n}\\{\frac {1}{(1-x)^{2}}}&=\sum _{n=1}^{\infty }nx^{n-1}\\{\frac {1}{(1-x)^{3}}}&=\sum _{n=2}^{\infty }{\frac {(n-1)n}{2}}x^{n-2}.\end{aligned}}}
 

{\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}&&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots &&{\text{for all }}x\\[6pt]\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}&&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots &&{\text{for all }}x\\[6pt]\tan x&=\sum _{n=1}^{\infty }{\frac {B_{2n}(-4)^{n}\left(1-4^{n}\right)}{(2n)!}}x^{2n-1}&&=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+\cdots &&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\sec x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}&&&&{\text{for }}|x|<{\frac {\pi }{2}}\\[6pt]\arcsin x&=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&&&{\text{for }}|x|\leq 1\\[6pt]\arccos x&={\frac {\pi }{2}}-\arcsin x\\&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}&&&&{\text{for }}|x|\leq 1\\[6pt]\arctan x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}&&&&{\text{for }}|x|\leq 1,\ x\neq \pm i\end{aligned}}}

Ratio test

The usual form of the test makes use of the limit
 

L=\lim _{{n\to \infty }}\left|{\frac {a_{{n+1}}}{a_{n}}}\right|.
L=\lim _{{n\to \infty }}\left|{\frac {a_{{n+1}}}{a_{n}}}\right|.
(1)
The ratio test states that:
  • if L < 1 then the series converges absolutely;
  • if L > 1 then the series is divergent;
  • if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

Root test

\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|},
converges then it equals C and may be used in the root test instead.
The root test states that:
  • if C < 1 then the series converges absolutely,
  • if C > 1 then the series diverges,
  • if C = 1 and the limit approaches strictly from above then the series diverges,
  • otherwise the test is inconclusive (the series may diverge, converge absolutely or converge conditionally).
There are some series for which C = 1 and the series converges, e.g.  

\textstyle \sum 1/{n^2}, and there are others for which C = 1 and the series diverges, e.g.  

\textstyle\sum 1/n.

Integral test for convergence

Consider an integer N and a non-negative, continuous function f defined on the unbounded interval [N, ∞), on which it is monotone decreasing. Then the infinite series
 

\sum _{{n=N}}^{\infty }f(n)

converges to a real number if and only if the improper integral
 

\int _{N}^{\infty }f(x)\,dx

is finite. In other words, if the integral diverges, then the series diverges as well.

Alternating series test

A series of the form
 

{\displaystyle \sum _{n=0}^{\infty }(-1)^{n}a_{n}=a_{0}-a_{1}+a_{2}-a_{3}+\cdots \!}

where either all an are positive or all an are negative, is called an alternating series.
The alternating series test then says: if  

a_{n} decreases monotonically and  

\lim _{{n\to \infty }}a_{n}=0 then the alternating series converges.