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. 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

**References**:**general form of a n-derivative***From:*Wiso <giurrerotipiacerebbe@hotmailtipiacerebbe.itipiacerebbe>

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