5
votes
Solving $2^a - 2^b = 2^x$ for extremely large $a$ and $b$
Assuming $a>b>0$:
$$
2^a-2^b=2^x\\
x=\frac{\ln(2^a-2^b)}{\ln2}
=\frac{\ln(2^b(2^{a-b}-1))}{\ln2}
=\frac{\ln(2^b)+\ln(2^{a-b}-1)}{\ln2}
=b+\frac{\ln(2^{a-b}-1)}{\ln2}\\
=b+\log_2(2^{a-b}-1)
$$
...
- 44.9k
4
votes
2
votes
Why can't NSolve find values that FindInstance can?
NSolve work if we use infinite precision number n,x,eps.
...
- 61.3k
2
votes
Accepted
How to replace==in the equation with ->?
A few observations
eq = 2 b == E^x0 - x0;
The Head of eq is ...
- 40.4k
1
vote
Finding zeroes of a modified Bessel function of the second kind
The problem is the fast decay of the absolute value as shown by
f[n_,ω_] := BesselK[I ω, 2 π n]
Plot[E^(6 + (ω^1.09)) f[ 2, ω], {ω, 6, 125}]
The exponent of $...
- 2,244
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