- #1
Zorba
- 77
- 0
In a past exam paper at my uni I am asked to show that the hermite polynomials are solutions of the hermite diff. equation but first there is the following
[tex]\Phi(x,t)=\exp (2xh-h^2)=\sum_{n=0}^{\infty} \frac{h^n}{n!}H_n(x)[/tex]
So I need to find the form of [tex]H_n[/tex] first, and I'm stuck. I tried writing
[tex]\exp (2xh-h^2)=\exp (2xh) \cdot \exp (-h^2)[/tex]
and then writing both of those as infinite series and using the Cauchy product formula but I couldn't get it to work out. I've tried a variety of other ways, but none seem to give me the proper formulas that I see on wikipedia and so on. So what the approach to take if you a priori know nothing about the Hermite polynomials, and you need to determine their form using the above definition?
[tex]\Phi(x,t)=\exp (2xh-h^2)=\sum_{n=0}^{\infty} \frac{h^n}{n!}H_n(x)[/tex]
So I need to find the form of [tex]H_n[/tex] first, and I'm stuck. I tried writing
[tex]\exp (2xh-h^2)=\exp (2xh) \cdot \exp (-h^2)[/tex]
and then writing both of those as infinite series and using the Cauchy product formula but I couldn't get it to work out. I've tried a variety of other ways, but none seem to give me the proper formulas that I see on wikipedia and so on. So what the approach to take if you a priori know nothing about the Hermite polynomials, and you need to determine their form using the above definition?
