Question about Darbaux Sums. I beg of ya, take a look for me.
So I feel like my answer must be wrong because it is too easy. I need the lower Darbaux sum for x^3 with partitions {0, .1, .4, 1}.
so, the \(m_i\) for the first partition would be 0, then .001 then .064, that's it?
but shouldn't there be 4?
@jim_thompson5910, @zepdrix, any ideas?
@dumbcow , I can't remember, have you covered this yet?
@ganeshie8 be my hero?
hey how is it different from riemann sum ?
it's the precursor to riemann. It's essentially the "left handed" sum from calc II, but Maybe it's just the class throwing me off here
technically it is defined as:\[L(P,f)=\sum^n_{i=1} m_i\Delta x_i\]\[m_i=inf~[f(x):x_{i-1}\le x\le x_i]\]
i can tell you the reason there are 3 m values is because with 4 points there are 3 intervals {x0,x1} [x1,x2] [x2,x3]
oh jeeze, yea, that makes sense
I can show you my work, the issue is I don't fully understand what the m_i are equal to.
m_i = inf M_i = sup in the ith interval
i got, m_i= 0,.001,.064
Hey I got the right answer!!!!
so I must have done it correctly! :D Thanks guys!
idk we both may be wrong, but based on the definition of sum it looks same as riemann sum for x^3 because min and max exist in each interval and they equal inf and sup
the def of riemann int has the upper and lower darbaux int. being equivalent, so maybe that is what it is
ohk.. do u get this for upper sum ? http://www.wolframalpha.com/input/?i=%280.1-0%29*0.1%5E3%2B%280.4-0.1%29*0.4%5E3+%2B+%281-0.4%29*1%5E3
wait, I haven't done it yet
give me a sec
Okay... i think there will be difference between riemann and darbaux when the curve is concave up or down. When that happens, the left and right values will not be inf and sup respectively always
I got .6193 for the upper
i got the same!
yay! Thanks, I was just doubting myself since it seemed so simple
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