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ND
Guest
that higher order derivatives follow pascal's triangle? It came up in an SGS trial i was doing today, pretty cool i recon.
redslert
yes, my actual brain
for those of us with less mathematically knowledge.....care to explain?
N
ND
Guest
Sorry i should've made the post more clear.
y=uv
y' = u'v+uv'
y'' = u''v + 2u'v' + uv''
y''' = u'''v + 3u''v' + 3u'v'' + uv'''
y'''' = u''''v + 4u'''v' + 6u''v'' + 4u'v''' + uv''''
Do those RHSs look familiar?
y=uv
y' = u'v+uv'
y'' = u''v + 2u'v' + uv''
y''' = u'''v + 3u''v' + 3u'v'' + uv'''
y'''' = u''''v + 4u'''v' + 6u''v'' + 4u'v''' + uv''''
Do those RHSs look familiar?
here's another relation:
1) find the first derivative f'(x) for the function f(x) from first principles
2) find the second derivative by the definition of the derivative
3) find the third derivate, using the definition of the derivative
etc.
the nth derivative should turn out to be
lim {h->0}[ SUM {k=0 to n} [(-1)^(n-k)]*(n C k)*f(x+kh) ]/h^n
1) find the first derivative f'(x) for the function f(x) from first principles
2) find the second derivative by the definition of the derivative
3) find the third derivate, using the definition of the derivative
etc.
the nth derivative should turn out to be
lim {h->0}[ SUM {k=0 to n} [(-1)^(n-k)]*(n C k)*f(x+kh) ]/h^n
redslert
yes, my actual brain
hmmmmm that is interesting! now i feel like a maths nerd 
what kinda question can they ask about that?
what kinda question can they ask about that?
