Re: general form of a n-derivative

• To: mathgroup at smc.vnet.net
• Subject: [mg71560] Re: [mg71542] general form of a n-derivative
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Thu, 23 Nov 2006 05:41:30 -0500 (EST)
• References: <200611221022.FAA04538@smc.vnet.net>

```On 22 Nov 2006, at 19:22, Wiso wrote:

> Hi, I have this function:
>
> \!\(f[x_] := Exp[\(-1\)/\((1 - x\^2)\)]\)
>
> and these:
>
> \!\(f\_n_[x_] := D[f[x], {x, n}]\)
>
> the n-derivative. I try to evaluate some of these:
> \!\(Table[FullSimplify[f\_i[x]], {i, 0, 5}]\  // TableForm\)
>
> now I want to infer the general form with Mathematica(I think
> somethink
> like:
>
> (-1)^(2 n) (Exp[1/(1-x^2)])^(2 n) P(x), where P(x) is a poly in x).
> Some
> ideas?
>

I believe the correct  answer is:

D[f[x], {x, n}] == (-1)^n*(E^(-1/(1 - x^2))/(x^2 - 1)^(2*n))*p[n][x]

where p[n][x] is a polynomial of degree 2n+1 with positive highest
coefficient.

function definition.

f[x_] := Exp[-(1 - x^2)^(-1)]

Here is the list of derivatives (I don't really want to show it here).

ls = Table[Together[D[f[x], {x, n}]], {n, 1, 20}];

Let us look at just one element:

ls[[3]]
-((4*x*(6*x^6 + 3*x^4 - 10*x^2 + 3))/(E^(1/(1 - x^2))*(x^2 - 1)^6))

As you see, we have a polynomial in the numerator and a product of an
exponential and a power of x^2 - 1 in the denominator. Let us first
look at just denominators:

ls2 = Denominator /@ ls
{E^(1/(1 - x^2))*(x^2 - 1)^2, E^(1/(1 - x^2))*(x^2 - 1)^4, E^(1/(1 -
x^2))*(x^2 - 1)^6, E^(1/(1 - x^2))*(x^2 - 1)^8, E^(1/(1 - x^2))*(x^2
- 1)^10, E^(1/(1 - x^2))*(x^2 - 1)^12, E^(1/(1 - x^2))*(x^2 - 1)^14,
E^(1/(1 - x^2))*(x^2 - 1)^16, E^(1/(1 - x^2))*(x^2 - 1)^18, E^(1/(1 -
x^2))*(x^2 - 1)^20, E^(1/(1 - x^2))*(x^2 - 1)^22, E^(1/(1 - x^2))*
(x^2 - 1)^24, E^(1/(1 - x^2))*(x^2 - 1)^26, E^(1/(1 - x^2))*(x^2 - 1)
^28, E^(1/(1 - x^2))*(x^2 - 1)^30, E^(1/(1 - x^2))*(x^2 - 1)^32, E^(1/
(1 - x^2))*(x^2 - 1)^34, E^(1/(1 - x^2))*(x^2 - 1)^36, E^(1/(1 - x^2))
*(x^2 - 1)^38, E^(1/(1 - x^2))*(x^2 - 1)^40}

It is easy to see that they have the form E^(1/(1 - x^2))*(x^2 - 1)^
(2n). Now let compute the polynomials in the numerator:

polys = Numerator /@ ls;

I really want to look at their highest terms:

ls1 = (With[{p = Exponent[#1, x]}, Coefficient[#1, x, p]*x^p] & ) /@
polys

{-2*x, 6*x^4, -24*x^7, 120*x^10, -720*x^13, 5040*x^16, -40320*x^19,
362880*x^22, -3628800*x^25, 39916800*x^28, -479001600*x^31,
6227020800*x^34, -87178291200*x^37, 1307674368000*x^40,
-20922789888000*x^43, 355687428096000*x^46, -6402373705728000*x^49,
121645100408832000*x^52, -2432902008176640000*x^55,
51090942171709440000*x^58}

So we see that they alternate in sign and have degrees equal to 3n+1.
This is the basis for our conjecture that the general term has the form

D[f[x], {x, n}] == (-1)^n*(E^(-1/(1 - x^2))/(x^2 - 1)^(2*n))*p[n][x]

where p[n][x] is a polynomial with positive leading coefficient of
degree 3n+1. Now we can prove this by induction.The proof I have got
is quite long, so I will only sketch it. The statement is clearly
true for n==1. Assume it is true for some n. Now compute:

v = ({Numerator[#1], Denominator[#1]} & )[Together[D[(-1)^n*(1/(E^(1/
(1 - x^2))*(x^2 - 1)^(2*n)))*p[n][x], x]]]

{(-1)^n*(Derivative[1][p*n][x]*x^4 - 4*n*(p*n)[x]*x^3 - 2*Derivative
[1][p*n][x]*x^2 + 4*n*(p*n)[x]*x - 2*(p*n)[x]*x + Derivative[1][p*n]
[x]), E^(1/(1 - x^2))*(x^2 - 1)^(2*n + 2)}

We immediately see that the denominator has the right form, and that
in the numerator we have a polynomial. Let's look at this polynomial
after dividing it by (-1)^(n+1):

Refine[Collect[(-1)^(n + 1)*v[[1]], x], Element[n,Integers]]

(-Derivative[1][p*n][x])*x^4 + 4*n*(p*n)[x]*x^3 + 2*Derivative[1][p*n]
[x]*x^2 - (4*n*(p*n)[x] - 2*(p*n)[x])*x - Derivative[1][p*n][x]

So the highest term will be the highest term of 4*n*(p*n)[x]*x^3-
Derivative[1][p*n][x]*x^4 provided this dos not vanish, and in that
case it will be of degree 3n+1 +3 == 3(n+1)+3. So it only remains to
show that the leading term of 4*n*(p*n)[x]*x^3-Derivative[1][p*n][x]
*x^4 has degree Degree[p[n]]+3 and has a positive coefficient. This
can again be done by induction starting with p[1], but I will leave
this part of the proof out out as I can't find a nice way to express
it in Mathematica.

It should not be to hard to determine the form of the polynomials p[n]
[x] exactly (rather than just their degree and the sigh of the
leading term) but i have not attempted to do so.

Andrzej Kozlowski
Tokyo, Japan

```

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