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

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mjc

Posts: 59
Registered: 9/22/05
Re: Cannot get inverse of y=x/logx to work
Posted: May 30, 2011 2:03 PM
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On May 30, 6:31 am, "G. A. Edgar" <ed...@math.ohio-state.edu.invalid>
wrote:
> In article <XKFEp.5313$aH5.5...@viwinnwfe02.internal.bigpond.com>, Brad
>
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>
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>
> 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.ohio-state.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)
?




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