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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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Brad Cooper

Posts: 171
Registered: 12/8/04
Cannot get inverse of y=x/logx to work
Posted: May 30, 2011 1:16 AM
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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.

(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.


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