In mathematics, the power series method is used to seek a power
series solution to certain differential equations. In general, such a
solution assumes a power series with unknown coefficients, and then
substitutes that solution into the differential equation to find a
recurrence relation for the coefficients. The power series method can be
applied to certain nonlinear differential equations, though with much
less flexibility.
Source Wikipedia.
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a_2 (x) f''(x) + a_1 (x) f'(x) + a_0 (x) f(x)` = 0
if a_2 is non zero for all value of x
f''(x) + (a_1 (x))/(a_2 (x)) f'(x) + (a_0 (x))/(a_2(x)) f` = 0
(where, `a_1/a_2` and `a_0/a_2 ` is an analytic function)
The power series method f = `sum_(k=0)^oo A_k x^k`
In calculus, the differentiation of x i.e., F'(x) = `sum_(n=0)^oo` an n(x - c)n - 1
Integration of x i.e., int` F(x)dx = `sum_(n=0)^oo` `a_n/(n+1)` (x - c)n + 1 + c
Power series method problem 1:
Find the power series representation of f(x) =` x/(x^2 + 36)` and determine the interval of convergence.
Solution:
x/(x^2 + 36)`
= ` x (1/(36+x^2))`
= `x/36 (1/(1+x^2/36)) `
= `x/36` ` (1/(1-(-x^2/36)))
= `x/36` `sum_(n=0)^oo` `(-x^2/36)^n
= `sum_(n=0)^oo` `(-1^n)(x^(2n+1)/36^(n+1))`
which convergence for `|-x^2/36|` < 1
i.e.,|x2 | < 36
i.e., |x| < 6 `
rArr` -6 < x < 6
check the end points,
For x = -6,
= sum_(n=0)^oo` `(-1^n)((-6)^(2n+1)/36^(n))`
= sum_(n=0)^oo` (-1)n (-6)
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and `lim_(n->oo)` (-1)n (6) does not exist. So,`sum` (-1)n (-6) diverges by the divergence test.
For x = 6,
= `sum_(n=0)^oo` `(-1^n)((6)^(2n+1)/36^(n))`
= `sum_(n=0)^oo` (-1)n (6)
which also diverges by the divergence test.
The interval of convergence is (-6, 6)
Source Wikipedia.
aving problem with The Harmonic Series keep reading my upcoming posts, i will try to help you.
Expression of Power series method:
the second order linear differential equation can be expressed as,a_2 (x) f''(x) + a_1 (x) f'(x) + a_0 (x) f(x)` = 0
if a_2 is non zero for all value of x
f''(x) + (a_1 (x))/(a_2 (x)) f'(x) + (a_0 (x))/(a_2(x)) f` = 0
(where, `a_1/a_2` and `a_0/a_2 ` is an analytic function)
The power series method f = `sum_(k=0)^oo A_k x^k`
In calculus, the differentiation of x i.e., F'(x) = `sum_(n=0)^oo` an n(x - c)n - 1
Integration of x i.e., int` F(x)dx = `sum_(n=0)^oo` `a_n/(n+1)` (x - c)n + 1 + c
Power series method problems:
Power series method problem 1:
Find the power series representation of f(x) =` x/(x^2 + 36)` and determine the interval of convergence.
Solution:
x/(x^2 + 36)`
= ` x (1/(36+x^2))`
= `x/36 (1/(1+x^2/36)) `
= `x/36` ` (1/(1-(-x^2/36)))
= `x/36` `sum_(n=0)^oo` `(-x^2/36)^n
= `sum_(n=0)^oo` `(-1^n)(x^(2n+1)/36^(n+1))`
which convergence for `|-x^2/36|` < 1
i.e.,|x2 | < 36
i.e., |x| < 6 `
rArr` -6 < x < 6
check the end points,
For x = -6,
= sum_(n=0)^oo` `(-1^n)((-6)^(2n+1)/36^(n))`
= sum_(n=0)^oo` (-1)n (-6)
I am planning to write more post on Pyramid Shape and samacheer kalvi books for 7th. Keep checking my blog.
and `lim_(n->oo)` (-1)n (6) does not exist. So,`sum` (-1)n (-6) diverges by the divergence test.
For x = 6,
= `sum_(n=0)^oo` `(-1^n)((6)^(2n+1)/36^(n))`
= `sum_(n=0)^oo` (-1)n (6)
which also diverges by the divergence test.
The interval of convergence is (-6, 6)
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