Math, asked by niharranjanpanda31, 5 days ago

Express 4x⁴+8x³+4x²-x-3 in terms of hermite polynomial​

Answers

Answered by chachathanos
1

Answer:

The Hermite polynomials H_n(x) are set of orthogonal polynomials over the domain (-infty,infty) with weighting function e^(-x^2), illustrated above for n=1, 2, 3, and 4. Hermite polynomials are implemented in the Wolfram Language as HermiteH[n, x].

The Hermite polynomial H_n(z) can be defined by the contour integral

H_n(z)=(n!)/(2pii)∮e^(-t^2+2tz)t^(-n-1)dt,

(1)

where the contour encloses the origin and is traversed in a counterclockwise direction (Arfken 1985, p. 416).

The first few Hermite polynomials are

H_0(x) = 1

(2)

H_1(x) = 2x

(3)

H_2(x) = 4x^2-2

(4)

H_3(x) = 8x^3-12x

(5)

H_4(x) = 16x^4-48x^2+12

(6)

H_5(x) = 32x^5-160x^3+120x

(7)

H_6(x) = 64x^6-480x^4+720x^2-120

(8)

H_7(x) = 128x^7-1344x^5+3360x^3-1680x

(9)

H_8(x) = 256x^8-3584x^6+13440x^4-13440x^2+1680

(10)

H_9(x) = 512x^9-9216x^7+48384x^5-80640x^3+30240x

(11)

H_(10)(x) = 1024x^(10)-23040x^8+161280x^6-403200x^4+302400x^2-30240.

(12)

When ordered from smallest to largest powers, the triangle of nonzero coefficients is 1; 2; -2, 4; -12, 8; 12, -48, 16; 120, -160, 32; ... (OEIS A059343).

The values H_n(0) may be called Hermite numbers.

The Hermite polynomials are a Sheffer sequence with

g(t) = e^(t^2/4)

(13)

f(t) = 1/2t

(14)

(Roman 1984, p. 30), giving the exponential generating function

exp(2xt-t^2)=sum_(n=0)^infty(H_n(x)t^n)/(n!).

(15)

Using a Taylor series shows that

H_n(x) = [(partial/(partialt))^nexp(2xt-t^2)]_(t=0)

(16)

= [e^(x^2)(partial/(partialt))^ne^(-(x-t)^2)]_(t=0).

(17)

Since partialf(x-t)/partialt=-partialf(x-t)/partialx,

H_n(x) = (-1)^ne^(x^2)[(partial/(partialx))^ne^(-(x-t)^2)]_(t=0)

(18)

= (-1)^ne^(x^2)(d^n)/(dx^n)e^(-x^2).

(19)

Now define operators

O^~_1 = -e^(x^2)d/(dx)e^(-x^2)

(20)

O^~_2 = e^(x^2/2)(x-d/(dx))e^(-x^2/2).

(21)

It follows that

O^~_1f = -e^(x^2)d/(dx)[fe^(-x^2)]

(22)

= 2xf-(df)/(dx)

(23)

O^~_2f = e^(x^2/2)(x-d/(dx))[fe^(-x^2/2)]

(24)

= xf+xf-(df)/(dx)

(25)

= 2xf-(df)/(dx),

(26)

so

O^~_1=O^~_2,

(27)

and

-e^(x^2)d/(dx)e^(-x^2)=e^(x^2/2)(x-d/(dx))e^(-x^2/2)

(28)

(Arfken 1985, p. 720), which means the following definitions are equivalent

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