Question

Solve Hermite's probabilistic equation given by L[u]=u''-xu=-nu

Answer 1

SOLUTION:

From given data,

Solve Hermite's probabilistic equation given by L[u]=u''-xu'=-nu

Let , L[u]=u''-x u'= - nu = > L[u]=u''-x u'+nu = 0

L[u]=u''-x u'+nu = 0 -------- (1)

Let ,

ke

kak.r

$u'' = \sum_{k=0}^{\infty } k(k-1) a_{k}x^{k-2}$

By substitution above all in equation (1)

$\sum_{k=0}^{\infty } k(k-1) a_{k}x^{k-2}$
-kak1
$+n\sum_{k=0}^{\infty } a_{k}x^{k}=0$

$\sum_{k=0}^{\infty } k(k-1) a_{k}x^{k-2}$$- x\sum_{k=0}^{\infty } k a_{k}x^{k}$

Σ nakrk

By replace

k = k-2

$\sum_{k=0}^{\infty } (k+2)*(k+1) * a_{k+2} * x^{k}$$+\sum_{k=0}^{\infty } (n-k)* a_{k}*x^{k}=0$

$\sum_{k=0}^{\infty }[ (k+1) *(k+2)* a_{k+2}+(n-k)a_{k}]x^{k} = 0$

Where,

$a_{k+2}$ = (k-n)*a_k / (k+1)(k+2)

Let us consider the recursion formula:

At k=0

$a_{2}$ = - n*a₀ / 1*2

At k=1

$a_{3}$ = (1-n)*a₁ / (1+1)(1+2)

$a_{3}$ =(1-n)*a₁ / 2*3

At k=2

$a_{4}$ = [ (2-n) /3*4 ] * [(-n) /2*1] * a₀

At k=3

$a_{5}$ = [ (3-n) /4*5 ] * [(1-n) /2*3] * a₁

Also ,

ke

= (a₀ + a₁x + a₂ x² + ...........)

= (a₀ - (na₀ / 2*1)* x²...........) + (a₁ x - ((1-n)a₁ / 2*3)* x³ +...........)

= (a₀ - (na₀ / 2*1)* x² - (n(2-n) / 4*3*2*1)* x⁴ +...........) + (a₁ x - ((1-n)a₁ / 2*3)* x³ + ((3-n)(1-n) / 5*4*3*2)* x⁵...........)

= (a₀*(1-n / 2*1)* x² - (n(2-n) / 4*3*2*1)* x⁴+..........) + (a₁ *(x - ((1-n)/ 2*3)* x³ + ((n-1)(n-3) / 5*4*3*2)* x⁵...........)

= (a₀*(1- n / 2!)* x² - (n(2-n) / 4!)* x⁴+..........) + (a₁ *(x - ((n-1)/ 3!)* x³ + ((n-1)(n-3) / 5!)* x⁵...........)

This is the solution

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