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

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