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Note that 2^1>1^32. :2^2
There are 3 solutions: 2 positive and 1 negative.
@АндрейВолков-ч4ы 2^0>0^322^-1
i just used log on both sides ran into w lambert func so i gave up ain't nobody got time to do bisection or Euler's method .
Don't give up! W-function is OK!
@@АндрейВолков-ч4ы motivated me to work on it again, setup newton's method f(m)=2^m−m^32, m_i+1 = m_i + f(m_i)/f'(m_i) got a solution 1.022 around when starting with a guess of 1 which wolfram alpha confirms ,using w-function is a lot more elegant.
2^m/32=mm/32*ln(2)=ln(m)ln(2)=ln(m)*32/mln(2)/32=ln(m)*m^(-1)ln(2)/32=ln(m)*e^ln(m^-1)-ln(2)/32=-ln(m)*e^-ln(m)W[-ln(2)/32]=-ln(m)-0,0221459=-ln(m)e^0,0221459=mm=1,02238m=256 por graph W-funciona.
(m^32)^1/32=|m| => 1 solution not found : m
Note that 2^1>1^32. :2^2
There are 3 solutions: 2 positive and 1 negative.
@АндрейВолков-ч4ы 2^0>0^32
2^-1
i just used log on both sides ran into w lambert func so i gave up ain't nobody got time to do bisection or Euler's method .
Don't give up! W-function is OK!
@@АндрейВолков-ч4ы motivated me to work on it again, setup newton's method f(m)=2^m−m^32, m_i+1 = m_i + f(m_i)/f'(m_i) got a solution 1.022 around when starting with a guess of 1 which wolfram alpha confirms ,using w-function is a lot more elegant.
2^m/32=m
m/32*ln(2)=ln(m)
ln(2)=ln(m)*32/m
ln(2)/32=ln(m)*m^(-1)
ln(2)/32=ln(m)*e^ln(m^-1)
-ln(2)/32=-ln(m)*e^-ln(m)
W[-ln(2)/32]=-ln(m)
-0,0221459=-ln(m)
e^0,0221459=m
m=1,02238
m=256 por graph W-funciona.
(m^32)^1/32=|m| => 1 solution not found : m