Derive Hermite Polynomial Generating Function from Recurrence Relation

[mathmajor314]

I am not sure how to format in LaTeX; I apologize for that.

The Hermite polynomials Hn(x) (physicist's version) satisfy
the recurrence relation,
H_{n+1}(x) - 2xHn(x) + 2nH_{n-1}(x) = 0; H0(x) = 1 and H1(x) = 2x:
Use this to derive the generating function for the Hermite polynomials, exp(2tx-t^2).

So far, I have gotten to Sum(H_{n+1}*exp(t)) = Sum{H_{n}[2x*exp(t)-2t*exp(t)]}.

I know I need to then integrate this then use H0(x) = 1 and H1(x) = 2x to solve for the constant of integration but I'm not sure how to do that.

 

[kurt.physics]

To solve for the generating function, start by multiplying both sides of the recurrence relation by an exponential factor exp(t): $$H_{n+1}(x) \cdot \exp(t) - 2xH_{n}(x)\cdot\exp(t) + 2nH_{n-1}(x)\cdot\exp(t) = 0$$Now, use the summation properties of the Hermite polynomials to rewrite the equation as: $$\sum_{n=0}^{\infty}H_{n+1}(x)\cdot\exp(t) - 2x\sum_{n=0}^{\infty}H_{n}(x)\cdot\exp(t) + 2\sum_{n=1}^{\infty}nH_{n-1}(x)\cdot\exp(t) = 0$$The first and second term can be combined and simplified to: $$\sum_{n=0}^{\infty}H_{n+1}(x)\cdot\exp(t)-2x\sum_{n=0}^{\infty}H_{n}(x)\cdot\exp(t) = \sum_{n=1}^{\infty}H_{n}(x)\cdot\exp(t) - 2x\sum_{n=0}^{\infty}H_{n}(x)\cdot\exp(t)$$Finally, use the initial conditions of the Hermite polynomials to get: $$\exp(t) -2x\sum_{n=0}^{\infty}H_{n}(x)\cdot\exp(t) = \sum_{n=1}^{\infty}H_{n}(x)\cdot\exp(t) -2xH_0(x)\cdot\exp(t)$$ $$\sum_{n=0}^{\infty}H_{n}(x)\cdot\exp(t) = \exp(2tx - t^2)$$Therefore, the generating function for the Hermite polynomials is $\exp(