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Topic:
Cannot get inverse of y=x/logx to work
Replies:
5
Last Post:
May 30, 2011 2:03 PM



mjc
Posts:
59
Registered:
9/22/05


Re: Cannot get inverse of y=x/logx to work
Posted:
May 30, 2011 2:03 PM


On May 30, 6:31 am, "G. A. Edgar" <ed...@math.ohiostate.edu.invalid> wrote: > In article <XKFEp.5313$aH5.5...@viwinnwfe02.internal.bigpond.com>, Brad > > > > > > > > > > Cooper <Brad.Cooper...@bigpond.com> wrote: > > Hi, > > > I cannot get the correct result for the inverse calculation of > > y = x/log(x). > > > On The fourth page (document pp. 240) of... > > > Partial Sums of Infinite Series, and How They Grow > > R. P. Boas, Jr. > > >http://mathdl.maa.org/images/upload_library/22/Ford/Boas.pdf > > > (It is 22 pages of pdf and takes a while to load) > > > The writer says... > > > The successive approximations to the inverse of y = x/logx are > > (if we write Ly for logy, L2y for log logy) > > > x = y*(Ly + L2y) + L2y/Ly + y*L2y/(Ly)^2*(1  1/2*L2y) + > > y*L2y/(Ly)^3*(1  3/2*L2y + 1/3*(L2y)^2) + > > y*L2y/(Ly)^4*(1  3*L2y + 11/6*(L2y)^2  1/4*(L2y)^3) > > > In the Computer Algebra Systsem MuPAD I wrote... > > > Inv := Y*(LY + L2Y) + L2Y/LY + Y*L2Y/(LY)^2*(1  1/2*L2Y) + > > Y*L2Y/(LY)^3*(1  3/2*L2Y + 1/3*(L2Y)^2) + > > Y*L2Y/(LY)^4*(1  3*L2Y + 11/6*(L2Y)^2  1/4*(L2Y)^3); > > > x := y > subs(Inv, Y = y, LY = ln(y), L2Y = ln(ln(y))); > > > This creates the inverse log function x = x(y) as shown in the paper. > > > Now, 1.295855509/ln(1.295855509) = 5.0 (approx.) > > > But in MuPAD I get x(1.295855509) = 3909.235291, not very close to 5.0 :) > > > I am interested to know if another CAS obtains a better result. > > > Am I missing something more basic in terms of the Mathematics? > > > Any help much appreciated. > > > Cheers, > > Brad > > Interesting question. Consider the function x/log(x) ... it decreases > for x in (0,e) and increases for x in (e,infinity). The minimum value > y=e occurs when x=e. Now for the part of the graph where x>e, Boas > has given an asymptotic expansion. For example, substitute y=5 into > the Boas series to get x=11.6, substitute x=11.6 into x/log(x) > to get 4.7, approximately right. > > But you are asking about the OTHER part of the graph, where x<e . > For that, the Boas series is nowhere near, as you noted. On that > part of the graph, the asymptotics look like > x ~= 1+1/y+(3/2)/y^2+(8/3)/y^3+(125/24)/y^4+(54/5)/y^5 > Using your values, substitute y=5 in there, and get 1.293 as you wanted. > > Others noted that the inverse is given in terms of the Lambert W > function: y = y*W(1/y) ... well, these two parts of the graph are > given as two branches of the W function. In Maple's notation, > y*LambertW(0,1/y) is the branch smaller than e, and > y*LambertW(1,1/y) is the branch larger than e. > >  > G. A. Edgar http://www.math.ohiostate.edu/~edgar/
Check your parentheses. It is not clear from what you have entered whether an expression written a/b*c should be a/(b*c) or (a/b)*c.
For example, is y*L2y/(Ly)^2*(1  1/2*L2y) y*L2y/((Ly)^2*(1  1/2*L2y)) or (y*L2y/(Ly)^2)*(1  1/2*L2y) ?



