Calculus the n'th derivative of $y_{n}=x^{n-1}e^{1/x}$ Without using Mathematical Induction to calculus the n'th derivative of the following function.
$y_{n}=x^{n-1}e^{1/x}$ , $n\in\mathbb{N}$
Find : $\frac{d^n}{dx^n}y_n$
I tried to finish the question from the given answer : $ y^{(n)}_{n}=\frac{(-1)^{n}e^{1/x}}{x^{n+1}} $
So, I want to know other method(s) to get the answer.
Thanks for your help!! :)
 A: Begin with
$${y_n}(x) = {x^{n - 1}}{e^{1/x}}$$
Then, taking the first derivative, we get
$${y_n}'(x)  = \left( {n - 1} \right){x^{n - 2}}{e^{1/x}} - {x^{n - 3}}{e^{1/x}}$$
This is
$$\tag{1}{y_n}'(x) = \left( {n - 1} \right){y_{n - 1}(x)} - {y_{n - 2}}(x) $$
You then get
$${y_n}''(x) = \left( {n - 1} \right)y{'_{n - 1}}(x) - y{'_{n - 2}}(x)$$
so you have to use $(1)$ to obtain that exclusively in terms of $y_n$.
You can do this for a couple of terms, conjecture a general formula, and prove it by induction. In fact, the formula will be only in terms of $n!/(n-k)!$ and $y_{n-k}$, it seems. When you get a general formula for $y_n^{(k)}$, you can plug in $n$ and get the formula you want.
Another option would be
$$\eqalign{
  & {y_n}' = \left( {n - 1} \right)\frac{1}{x}{x^{n - 1}}{e^{1/x}} - \frac{1}{{{x^2}}}{x^{n - 1}}{e^{1/x}}  \cr 
  & {y_n}' = \left( {n - 1} \right)\frac{1}{x}{y_n} - \frac{1}{{{x^2}}}{y_n}  \cr 
  & {y_n}^\prime (x) = \left( {n - 1} \right)\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right){y_n} \cr} $$
Then
$${y_n}''(x) = \left( {n - 1} \right)\left( { - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}}} \right){y_n} + \left( {n - 1} \right)\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)y{'_n}$$
So
$${y_n}''(x) = \left( {n - 1} \right)\left( { - \frac{1}{{{x^2}}} + \frac{2}{{{x^3}}}} \right){y_n} + {\left( {n - 1} \right)^2}{\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)^2}{y_n}$$
But it seems much more complicated, due to the repeated use of the product rule and chain rule involved.
A: I don't know if you got the answer you need. Anyway, I tried the Laurent expansion and it worked
Firstly, we have some conclusion
$$\frac{d^{n}}{dx^{n}}x^{-m}=\frac{(-1)^{n}(m+n-1)!x^{-m-n}}{(m-1)!}$$ if m is positive integer
And
$$\frac{d^{n}}{dx^{n}}x^{m}=0$$ if m is non-negative integer and m<n
We have
$$e^{\frac{1}{x}}=\sum_{i=0}^\infty\frac{x^{-i}}{i!}$$
So, $$y_n=\sum_{i=0}^\infty\frac{x^{n-i-1}}{i!}$$
Now, n-i-1 is always less than n, so
$$\frac{d^n}{dx^n}\sum_{i=0}^{n-1}\frac{x^{n-i-1}}{i!}=0$$
We then get
$$y_{n}^{(n)}=\sum_{i=n}^\infty\frac{\frac{(-1)^n(-n+i+1+n-1)!}{(-n+i)!}x^{-i-1}}{i!}$$
$$=\sum_{i=n}^\infty\frac{(-1)^nx^{-i-1}}{(i-n)!}$$
Now let k=i-n, then we have
$$y_{n}^{(n)}=\sum_{k=0}^\infty\frac{(-1)^nx^{-n-k-1}}{k!}$$
$$=\frac{(-1)^n}{x^{n+1}}\sum_{k=0}^\infty\frac{x^{-k}}{k!}$$
This is $$y_{n}^{(n)}=\frac{(-1)^ne^{\frac{1}{x}}}{x^{n+1}}$$
Hope this help
