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$e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}$

$\frac{\partial}{\partial t }(e^{z\sqrt{1-t}})=\frac{\partial}{\partial t }(\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!})$

$\frac{-z}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=1}^\infty \frac{u_k(z)t^{k-1}}{{(k-1)}!}=\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)t^{k}}{{k}!}$

$\frac{-1}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)}{z}\frac{t^{k}}{{k}!}$

$\frac{\partial}{\partial z }(\frac{-1}{2\sqrt{1-t} }e^{z\sqrt{1-t}})=\frac{\partial}{\partial z }(\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)}{z}\frac{t^{k}}{{k}!})$

$\frac{-e^{z\sqrt{1-t}}}{2}=\sum \limits_{k=0}^\infty \frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})\frac{t^{k}}{{k}!}$

$\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}=-2\sum \limits_{k=0}^\infty \frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})\frac{t^{k}}{{k}!}$

$u_k(z)=-2\frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})$

for $t=0$,$u_0(z)=e^z$

How to find the general formula of $u_k(z)$ ? I would like to learn the methods to solve such differential equations.

Thanks a lot for answers.

EDIT:

We can find $u_1(z)$ as shown below

$\frac{\partial}{\partial t }(e^{z\sqrt{1-t}})=\frac{-z}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=1}^\infty \frac{u_k(z)t^{k-1}}{{(k-1)}!}$

for $t=0$, $u_1(z)=\cfrac{-ze^{z}}{2} $

If we continue to derivate in such way, we can find all $u_k(z)$ but it seems long method. I am looking for easier method.

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    @GEdgar We have $e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}$2012-10-16

2 Answers 2

1

A typical approach for this kind of problems (Hermite, Legendre, etc) is to obtain recurrence relations.

You've proven that $ u_n(z) = - 2 \frac{d}{d z} \left(\frac{u_{n+1}(z)}{z}\right). $ A more simple relation can be derived: $ \frac{d^2}{d z^2} e^{z\sqrt{1-t}} = (1-t) e^{z \sqrt{1-t}} = (1-t) \sum_0^\infty \frac{u_n(z)}{n!}t^n = \sum_0^\infty \frac{u_k''(z)}{n!}t^n $ Arranging orders of $t$ and using your relation \begin{align} u_0'' - u_0 &= 0, \qquad k = 0,\\ u_n'' - u_n + n u_{n-1} &= 0, \qquad n \ge 1, \end{align} then $u_0^{(1)} = e^z$ and $u_0^{(2)} = e^{-z}$, and the ode $ u_n'' - \frac{2 n}{z} u_n' - \left(1 - \frac{2 n}{z^2}\right)u_n = 0, \qquad n \ge 1. $ Taking the change of variables $ u_n(z) = z^{n + 1} v_n(z) $ we have $ v_n'' + \frac{2}{z} v_n' - \big(n(n-1) + z^2\big) v_n = 0, \qquad n \ge 1. $ hence $ u_n^{(1)}(z) = z^{n + 1} i_{-n}(z), \quad u_n^{(2)}(z) = z^{n + 1} k_{-n}(z), $ where $i_n$ and $k_n$ are the Modified Spherical Bessel Functions of the first and second kind.

2

Using the exponential and binomial series, $ e^{z\sqrt{1-t}} =\sum_{j=0}^\infty\frac{z^j(1-t)^{j/2}}{j!} =\sum_{k=0}^\infty\frac{t^k}{k!}\sum_{j=0}^\infty{j/2\choose k}\frac{(-1)^kk!z^j}{j!} $

Performing the sum over $j$ gives the $u_j(z)$ in terms of modified Bessel functions of the first kind: $ u_j(z) =\sum_{j=0}^\infty{j/2\choose k}\frac{(-1)^kk!z^j}{j!} =\left(\pi\frac{z}{2}\right)^{\frac{1}{2}}\left(-\frac{z}{2}\right)^jI_{1/2-j}(z) $

I'm not sure how to solve your differential equation, but I think trying to relate it to differential equations of Bessel functions might lead somewhere.