5 votes

Solving $2^a - 2^b = 2^x$ for extremely large $a$ and $b$

Assuming $a>b>0$: $$ 2^a-2^b=2^x\\ x=\frac{\ln(2^a-2^b)}{\ln2} =\frac{\ln(2^b(2^{a-b}-1))}{\ln2} =\frac{\ln(2^b)+\ln(2^{a-b}-1)}{\ln2} =b+\frac{\ln(2^{a-b}-1)}{\ln2}\\ =b+\log_2(2^{a-b}-1) $$ ...
Roman's user avatar
  • 44.9k
4 votes

Why can't NSolve find values that FindInstance can?

Use FindRoot ...
Bob Hanlon's user avatar
  • 149k
2 votes

Why can't NSolve find values that FindInstance can?

NSolve work if we use infinite precision number n,x,eps. ...
cvgmt's user avatar
  • 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 ...
eldo's user avatar
  • 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 $...
Roland F's user avatar
  • 2,244

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