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About LinkBacksn*2^n = 2^53
HOw to solve for n??
thank you
Formula iteration will solve this with a bit of cunning:Originally Posted by tea217
Rewrite the equation as:
n=ln[(2^55)/n]/log(2)
Then iterate:
n(r+1)=ln[(2^55)/n(r)]/log(2)
where n(r) is the r-th iterate.
Then with n(0)=1, we get:
n(1)=53,
n(2)=47.2721,
n(3)=47.4371,
n(4)=47.4321
n(5)=47.4322
n(6)=47.4322
With n=47.4322 the two sides of the original equation agree to 5 significant digits.
RonL
This can also be solved with the LamberW function.
If A=w*e^w, then W=LambertW(A).
Your equation may be written:
n* e^(ln(2) n)=2^53,
or:
[ln(2)*n] e^[ln(2)*n]=ln(2)*2^53,
so:
ln(2)*n=LambertW(ln(2)*2^53),
or:
n=LambertW(ln(2)*2^53)/ln(2).
Now LambertW(ln(2)*2^53)~=32.8775, so n=32.8775/ln(2) ~=47.4322.
RonL
"n=ln[(2^55)/n]/log(2)"
how did u get this statement?
why is there a ln and log? what base is log(2) ?
this method seems easier to understand than the other one, which i have never used that function before.
But i really appricate your help.
Typo, default base for log is e, so when I forget to type ln and type log
I usually mean log base e.
We have:
n 2^n = 2^53
which we may rewrite as:
n e^(n ln(2))=2^53
so:
e^(n ln(2))=(2^53)/n
Now taking logs (ln's) gives:
n ln(2)= ln((2^53)/n)
n= ln((2^53)/n)/ln(2)
The equation is reorganised in this manner so that formula iteration
will converge (with the obvious rearrangement iteration diverges)
RonL
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