Of late, I have come to the conclusion that I never really learned integration by parts when I was studying calculus. Or perhaps that I learned it by rote, rather than by understanding. So I recently took steps to correct that....
∫ uv' = uv - ∫ u'v
That's what I remember; and the tricky part was always to pick the right u's and v's and to keep track of everything as you go. But where does it come from? Take the derivative of both sides:
d/dx[∫ uv' ] = d/dx[ uv - ∫ u'v ]
d/dx[∫ uv' ] = d/dx[ uv ] - d/dx [ ∫ u'v ]
d/dx[∫ uv' ] + d/dx [ ∫ u'v ] = d/dx[ uv ]
uv' + u'v = d/dx[ uv ]
And hey presto! you have the product rule. Don't I feel silly.
Working backwards now, I tried to rediscover the proof of the product rule.
d/dx [ uv ]
= lim [uv(x+h)-uv(x) ] /h
= lim [u(x+h)v(x+h)-u(x)v(x)] / h
= lim [u(x+h)v(x+h)-u(x)v(x+h)+u(x)v(x+h)-u(x)v(x)] / h
= lim [u(x+h)v(x+h)-u(x)v(x+h)]/h + [u(x)v(x+h)-u(x)v(x)]/h
= lim [u(x+h)v(x+h)-u(x)v(x+h)]/h + lim [u(x)v(x+h)-u(x)v(x)]/h
= lim [v(x+h)[u(x+h)-u(x)]/h] + u(x)lim [v(x+h)-v(x)]/h
= lim v(x+h) lim [u(x+h)-u(x)]/h] + u(x)lim [v(x+h)-v(x)]/h
= v(x)u'(x) + u(x)v'(x)
= vu' + uv'
But I palmed a card when evaluating the limit in the first term. Is the product of the limits really the limit of the products? Of course it is, but what was the demonstration?
What we are trying to show is that for any given distance ε, there exists a distance δ that ensures that for all h < δ it will be true that | AB(x+h) - L(A)L(B) | < ε.
Definition: L(A) = lim A(x+h) as h->0
AB(x+h) - L(A)L(B)
= A(x+h)B(x+h) - L(A)L(B)
= A(x+h)B(x+h) - L(A)B(x+h) + L(A)B(x+h) - L(A)L(B)
= [A(x+h)-L(A)]B(x+h) + L(A)[B(x+h)-L(B)]
< [ε](ε+L(B))+L(A)[ε)]
< ε^2 + ε L(B) + ε L(A)
So we have an expression for the uncertainty in the product, given the uncertainties in A and B. Given a target uncertainty in the product, we choose a value for delta such that the sum above is less than ε. That is, by the definition of L(A) and L(B), we can always find δ so that the uncertainty is less than ε / (1+L(A)+L(B)); therefore the uncertainty of the product is less than ε and the limit is achieved.
Confused? Not to worry: Karl Hahn's explanation is clearer.
February 5, 2005 8:59 AM
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