How do I evaluate several n-th derivatives of a function at one point?

Question

I have a question where I have to compute a Table containing $f^{(n)}(0)$ for n = 1, ..., 5, where $f^{(n)}$ denotes the $n$th derivative of $f$.

For the function $f(x)=x\mathrm{e}^{-x}$, I have:

f[x_]:=xe^-x
Table[D[f[x],{x,n}],{n,1,5}]//TableForm

How can I solve for the value of each derivative at 0?

Answer 1

Since Vitaliy already answered the question, I'll just add another answer to confuse you. To get the general form of the $n$-th derivative, you could use the properties of the Taylor series as follows:

Clear[n];
c[n_] = FullSimplify[SeriesCoefficient[n! x Exp[-x], {x, 0, n}], 
  n >= 0]

$\begin{cases} -(-1)^n n & n\geq 1 \\ 0 & \text{True} \end{cases}$

This is the $n$-th coefficient in the series expansion of your function around $x = 0$ (with the factorial $n!$ canceled). It is the thing you're supposed to tabulate. To verify this, here is the table explicitly:

Array[c, {5}] // TableForm

$\begin{array}{c} 1 \\ -2 \\ 3 \\ -4 \\ 5 \\\end{array}$

The nice thing about c[n] is of course that you no longer need to make such tables if you have a closed form for the derivatives. If you do want to make tables, though, then c[n] allows you to do this at very high speed because the definition c[n_] = ... using Set (the = sign) caused the right-hand side to be evaluated once and for all, so it doesn't have to be repeated every time you change n.

Answer 2

You have 2 typos. No space in xe and use capital E not e - or better Exp[x]

You have nicely behaving functions, so this would work:

f[x_] := x Exp[-x]
Table[D[f[x], {x, n}] /. x -> 0, {n, 1, 5}]

{1, -2, 3, -4, 5}

But this is dangerous and it is better to use this:

f[x_] := x Exp[-x]
Table[Limit[D[f[x], {x, n}], x -> 0], {n, 1, 5}]

{1, -2, 3, -4, 5}

To illustrate, imagine you have other function:

f[x_] := SinIntegral[x]
Table[D[f[x], {x, n}], {n, 1, 5}] // FullSimplify // Column

Using /.x- > 0 will now give error due to x in denominator, while Limit is fine:

f[x_] := SinIntegral[x]
Table[Limit[D[f[x], {x, n}], x -> 0], {n, 1, 5}]

{1, 0, -(1/3), 0, 1/5}

Answer 3

There is another approach that sometimes works better (gives closed-form expressions rather than recurrence relations):

In[1]:= InverseFourierTransform[(-I k)^n FourierTransform[1/(1 + x^2)^Log[2], x, k] , k, x]
Out[1]= (2^(-1 + n - 1/2 Log[1/x^2])
      Abs[x]^-Log[2] ((-I)^
      n ((1 + n) x Gamma[(1 + n)/2] Gamma[
      n/2 + Log[2]] Hypergeometric2F1[(1 + n)/2, n/2 + Log[2], 1/
      2, -x^2] (n + Log[4]) - 
    2 I Gamma[1 + n/2] Gamma[
      1/2 (1 + n + Log[4])] ((1 + x^2) Hypergeometric2F1[(2 + n)/
         2, 1/2 (1 + n + Log[4]), -(1/2), -x^2] - 
       Hypergeometric2F1[(2 + n)/2, 1/2 (1 + n + Log[4]), 1/
         2, -x^2] (1 + x^2 (3 + 2 n + Log[4])))) + 
 I^n ((1 + n) x Gamma[(1 + n)/2] Gamma[
      n/2 + Log[2]] Hypergeometric2F1[(1 + n)/2, n/2 + Log[2], 1/
      2, -x^2] (n + Log[4]) + 
    2 I Gamma[1 + n/2] Gamma[
      1/2 (1 + n + Log[4])] ((1 + x^2) Hypergeometric2F1[(2 + n)/
         2, 1/2 (1 + n + Log[4]), -(1/2), -x^2] - 
       Hypergeometric2F1[(2 + n)/2, 1/2 (1 + n + Log[4]), 1/
         2, -x^2] (1 + x^2 (3 + 2 n + Log[4]))))))/((1 + n) 
       Sqrt[Pi] x Gamma[Log[2]] (n + Log[4]))

It also can be used to find repeated anti-derivatives.