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Re: general form of a n-derivative
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71562] Re: [mg71542] general form of a n-derivative
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Thu, 23 Nov 2006 05:41:32 -0500 (EST)
*References*: <200611221022.FAA04538@smc.vnet.net> <FB3C4201-33DA-4157-B95F-13263EA744C0@mimuw.edu.pl>
On 23 Nov 2006, at 10:37, Andrzej Kozlowski wrote:
>
> 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.
>
>
> First, let us investigate this empirically. With start with the
> 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
>
I noticed two things. First, in the formulas above p*n should
everywhere be replaced by p[n]. I had p[n] correctly in my
Mathematica notebook, but somehow it was converted to p*n by the
CopyAsInpuTForm command in Ingolf Dahl's SetFaceAndFont2 palette that
I have been using to avoid problems with Mathematica formulas pasted
into my Mail program. It seems that there is a bug in the palette.
All the p*n above should be replaced by p[n] - which means a
polynomial of degree 3n+1 in the formula for th en-th derivative and
not a product of p and n.
I also just noticed that the remaining part of the argument can also
be done in an almost trivial way, so I have decided to include it
after all. It remains to show that
4*n*(p[n])[x]*x^3-Derivative[1][p[n]][x]*x^4
has a non-vanishing highest term with positive coeffcient. Suppose
then p[n_]:=a[n] x^(3n+1) + ... where a[n]>0. Without loss of
generality can sssume that
p[n_][x_] = a[n] x^(3n + 1);
Now
Simplify[4*n*p[n][x]*x^3 - Derivative[1][p[n]][x]* x^4]
(n - 1)*x^(3*n + 4)*a[n]
and clear (n-1)*a[n] is the highest term and is non zero. This
actually also tells us what the highest terms are: they are just
successive factorials. So making use of this and correcting some
errors in the indices above we get the following answer:
D[f[x],{x,n}]==(-1)^n*(E^(-1/(1 - x^2))/(x^2 - 1)^(2*n))*p[n][x]
where p[n][x]= (n+1)! x^(3n-2) + lower degree terms.
Indeed, you can check that
In[37]:=
Table[(-1)^n*(n + 1)!*x^(3*n - 2), {n, 1, 20}]
Out[37]=
{-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}
which is exactly ls1 above.
Andrzej Kozlowski
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