How does one show that $f(\xi)$ will behave as $e^{\xi^2}$ if $\lambda<0$ in the Hermite differential equation $\frac{d^2f}{d\xi^2}-2\xi\frac{df}{d\xi}+2\lambda f(\xi)=0$
This problem comes in solving the linearly independent Schrodinger equation correpsonds to a 1D harmonic oscillator.
It says since $f(\xi)$ behaves as $e^{\xi^2}$ if $\lambda<0$, $f(\xi)\to +\infty$ as $x\to \pm \infty$, the corresponding solutions are physically not desirable, thus $\lambda \geq0$.
how do I prove that $f(\xi)$ behaves as $e^{\xi^2}$ if $\lambda<0$ mathematically ?
Note: I understand how to solve the Hermite differential equation when $\lambda\geq 0$ to obtain Hermite functions of different order using the Power series method.
Reference: Page 74 of ONE DIMENSIONAL HARMONIC OSCILLATOR
Possible Way
As I have seen in some reference,
Making use of the Frobenius method, taking $f(\xi)=\xi^s\sum_{r=0}^\infty a_r\xi^r$
$$ \frac{a_{r+2}}{a_r}=\frac{2(r+s-\lambda)}{(r+s+2)(r+s+1)}=\frac{2(1+\frac{s}{r}-\frac{\lambda}{r})}{r(1+\frac{s+2}{r})(1+\frac{s+1}{r})}\bigg|_{r\to\infty}\to \frac{2}{r} $$
Consider the series $e^{\xi^2}=1+\xi^2+\frac{\xi^4}{2!}+\cdots$. The ratio of the adjacent terms is $$ \frac{b_{r+2}}{b_r}=\frac{(r/2)!}{(\frac{r+2}{2})!}=\frac{\bigg(\dfrac{r}{2}\bigg)!}{\bigg(\dfrac{r+2}{2}\bigg)!}=\frac{1}{\frac{r}{2}+1}\bigg|_{r\to\infty}\to \frac{2}{r} $$ asymptotically, $f(\xi)$ is behaving like $e^{\xi^2}$, thus $\psi=e^{-\xi^2/2}f(\xi)$ goes like $e^{+\xi^2/2}$.
Is it a well defined explanation ?
The long story short is that the exponential approximation is only valid if $\boldsymbol{|\lambda|}$ is close to $\boldsymbol1$. The explanation is below.
Putting $\lambda\to -\lambda$ into the equation we have $$y''-2xy'-2\lambda y=0$$ The solution to this equation can be obtained via a power series approach. We can write $$y(x;\lambda)=c_1\cdot M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)+ c_2\cdot 2^{-\lambda}U\left(\frac{-\lambda}{2};\frac{1}{2}~\bigg|~x^{2}\right) $$ These are the confluent hypergeometric functions of the first and second kind. The strange delimiters in the arguments is a stylistic choice due to their relation to the generalized hypergeometric function. IF $y$ satisfies the convenient conditions $y(0)=1$, $y'(0)=0$ then we can discard the confluent hypergeometric of the second kind and write $$y(x;\lambda)= M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)$$ $M$ is defined as $$M(a;b~|~z):=\sum_{k=0}^\infty \left[\prod_{i=0}^{k-1}\frac{a+i}{b+i}\frac{x^k}{k!}\right]$$ Which would mean that $$M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)=\sum_{k=0}^\infty \left[\prod_{i=0}^{k-1}\frac{\lambda/2+i}{1/2+i}\frac{x^{2k}}{k!}\right]=\sum_{k=0}^\infty \left[\prod_{i=0}^{k-1}\frac{1+2i/\lambda}{1+2i}\frac{(\sqrt{\lambda}~x)^{2k}}{k!}\right]$$ Assuming $\lambda$ is close to $1$ the product vanishes and we have $$y(x;\lambda)=M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)\approx\sum_{k=0}^\infty \frac{(\sqrt{\lambda}~x)^{2k}}{k!}=\exp\left(\lambda x^2\right).$$
This is OK, but we can think of pathological examples that don't follow this reasoning. Consider the two functions $e^{z}=\sum_{n=0}^\infty a_n z^n$ and $\sin(z)=\sum_{n=0}^\infty b_nz^n$ where $a_n=\frac{1}{n!}$ and $$b_{n} =\begin{cases} 0 & n=2k\\ \frac{1}{n!} & n=4k+1\\ \frac{-1}{n!} & n=4k+3 \end{cases}$$ It is true that for odd $n$, $$\frac{a_{n+4}}{a_n}=\frac{b_{n+4}}{b_n}=\frac{1}{(n+1)(n+2)(n+3)(n+4)}$$ But I think it would be foolish to say $\sin(x)\to e^x$ for $x\to\infty$.