Upper Lower Sum

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March 10th 2011, 06:16 PM
mathematic

Upper Lower Sum

Consider f(x)=2x+1 over [1,3]. Let P be the partition consisting of points {1,3/2,2,3}. Find L(f,P), U(f,P).
I'm having trouble calculating. I have L(f,P)=17/2 and U(f,P)=23/2 but don't know how to get there.
L(f,P)=sum(inf f(x)*2)
U(f,P)=sum(sup f(x)*2)
March 10th 2011, 06:21 PM
Drexel28

Quote:

Originally Posted by mathematic

Consider f(x)=2x+1 over [1,3]. Let P be the partition consisting of points {1,3/2,2,3}. Find L(f,P), U(f,P).
I'm having trouble calculating. I have L(f,P)=17/2 and U(f,P)=23/2 but don't know how to get there.
L(f,P)=sum(inf f(x)*2)
U(f,P)=sum(sup f(x)*2)

Recall that in general for a partition $\displaystyle P:a=x_0\leqslant\cdots\leqslant x_n=b$ one has that $\displaystyle U(P,f)=\sum_{j=1}^{n}\sup_{x\in[x_{j-1},x_j]}f(x)\Delta_x_j$ . So, the only possible ambiguity here is how to compute $\displaystyle \sup_{x\in[x_{j-1},x_j]}f(x)$ but since your $f$ is strictly increasing this is just $f(x_j)$ . Make sense?
March 10th 2011, 06:43 PM
mathematic

so the xj is 3 right?
so f(3)=2(3)+1=17(2)
L(f,p)=34?
U(f,p)=f(1)(2)=[2(1)+1]2=3*2=6
March 10th 2011, 06:50 PM
Drexel28

Quote:

Originally Posted by mathematic

so the xj is 3 right?
so f(3)=2(3)+1=17(2)
L(f,p)=34?
U(f,p)=f(1)(2)=[2(1)+1]2=3*2=6

No. Look again at the definition. We have $\underbrace{1}_{x_0}\leqslant \underbrace{\frac{3}{2}}_{x_1}\leqslant \underbrace{2}_{x_2}\leqslant \underbrace{3}_{x_3}$ .
March 10th 2011, 06:57 PM
mathematic

So f(1)*2+f(3/2)*2+f(2)*2+f(3)*3
=6+8+5+17
March 10th 2011, 07:05 PM
Drexel28

Quote:

Originally Posted by mathematic

So f(1)*2+f(3/2)*2+f(2)*2+f(3)*3
=6+8+5+17

Why are you multiplying by the things you did? Look again at the definition of the upper sum. You should, for example, by multiplying $f(1)$ by $\frac{3}{2}-1=\frac{1}{2}$
March 10th 2011, 07:17 PM
mathematic

so f(1)(1/2)+f(3/2)(2-3/2)+f(2)(3-2)
3/2+4(1/2)+5
3/2+4/2+5
7/2+10/2=17/2

f(3/2)(1/2)+f(2)(2-3/2)+f(3)(3-2)
4/2+5(1/2)+7
2+5/2+7
9/2+14/2
23/2