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+--- Thread: using cosh(x) ? (/showthread.php?tid=718)
using cosh(x) ? - tommy1729 - 11/19/2011
no , im not kidding , apart from my sinh method im considering a cosh method.
although it might be more complicated or speculative.
it might be insightfull to plot things.
consider
exp(1/e)<b<sqrt(e)
f(x) - f^[-1](x) = 2*cosh(ln(b)*x)
for variable x and fixed b ( base ).
i bet you can find more functional equations from this one.
for instance to express f^[-1](x) in terms of f(x) and elementary functions.
the solution is unique for R-> R.
replace x by f(x) and let g(x) be the super of f(x).
then we get a difference equation for g(x).
we solve that difference equation , now we go from g(x) to f(x) or we find the taylor for f(x) from the equation ( or both , in programming ).
similar to the sinh method : since f is close to 2*cosh(ln(b)*x)) and 2*cosh(ln(b)*x) is close to b^x , we have a new method.
RE: using cosh(x) ? - tommy1729 - 11/20/2011
also , we finally have boundaries.
for large x , f^[1/2](x) + f^[1/2](ln_b(x)) is a good bound for tet(b,1/2,x).
RE: using cosh(x) ? - tommy1729 - 11/20/2011
hmm
i was thinking about
f(x) - f[-1](x) = 2*sinh(ln(b)*x).
for approximating (b^x)^[theta] for bases b with eta < b < sqrt(e) and x > e^e.
RE: using cosh(x) ? - tommy1729 - 11/20/2011
a possible improvement would be
f(x) - f^[-1](x) = 2*sinh(ln(b)*x) + (e-2)*x/e.
f(0) = 0
tommy1729
RE: using cosh(x) ? - sheldonison - 11/20/2011
(11/19/2011, 11:56 PM)tommy1729 Wrote: no , im not kidding , apart from my sinh method im considering a cosh method.What is the fixed point for this method? The fixed point for 2sinh was zero.
RE: using cosh(x) ? - tommy1729 - 11/20/2011
(11/20/2011, 03:18 PM)sheldonison Wrote:(11/19/2011, 11:56 PM)tommy1729 Wrote: no , im not kidding , apart from my sinh method im considering a cosh method.What is the fixed point for this method? The fixed point for 2sinh was zero.
the fixed point is pi/2 i for the cosh method.
for the ' new ' sinh method it is also 0.